PHYS 232 - UNIVERSITY PHYSICS
II - Temperature and heat
Question Bank - Set 10
Liberty University
Question 1
Question
A copper block with a mass of 500 g is initially at a temperature of 100
°
C. It is
placed in a container filled with 1 kg of water at 20
°
C. Assuming no heat is lost
to the surroundings, calculate the final equilibrium temperature of the system.
Given: Specific heat capacity of copper = 0.385 J/g
°
C Specific heat capacity
of water = 4.18 J/g
°
C Heat of fusion of copper = 205 J/g Heat of vaporization
of water = 2260 J/g
Solution
Step 1: Calculate the heat lost by the copper block as it cools down from 100
°
C
to the final equilibrium temperature.
The formula to calculate heat energy is:
Q=mc∆T
where: Q= heat energy (in joules) m= mass of the object (in grams) c=
specific heat capacity (in joules/gram
°
C) ∆T= change in temperature (in
°
C)
Substitute the values for copper: m= 500 g c= 0.385 J/g
°
C ∆T= 100 −Tf
(where Tfis the final equilibrium temperature)
The heat lost by the copper block can be calculated as:
Qcopper = 500 ×0.385 ×(100 −Tf)
Step 2: Calculate the heat gained by the water as it warms up from 20
°
C to
the final equilibrium temperature.
Substitute the values for water: m= 1000 g c= 4.18 J/g
°
C ∆T=Tf−20
The heat gained by the water can be calculated as:
Qwater = 1000 ×4.18 ×(Tf−20)
Step 3: Set up the equation and solve for the final equilibrium temperature,
Tf.
Since the system is isolated and there is no heat loss to the surroundings:
Qcopper =Qwater
So, we have:
500 ×0.385 ×(100 −Tf) = 1000 ×4.18 ×(Tf−20)
Solve this equation to find the final equilibrium temperature, Tf.
Question 2
Question
A metal rod of length 1.5 m and diameter 2 cm is initially at a temperature of
100
°
C. It is then placed in a large pool of water at 20
°
C. If the heat capacity of
the metal is 450 J/kg
°
C and the thermal conductivity is 70 W/m
°
C, calculate
the time it takes for the center of the rod to reach a temperature of 40
°
C. Assume
the surroundings remain at a constant temperature of 20
°
C.
Solution
Step 1: Find the initial temperature difference between the center of the rod
and the surroundings. Given: Initial temperature of the rod, T0= 100CTem-
perature of the surroundings, Ts= 20C
The initial temperature difference, ∆T0=T0−Ts= 100C−20C= 80C
Step 2: Find the final temperature difference between the center of the rod
and the surroundings. The final temperature of the rod, Tf= 40C
The final temperature difference, ∆Tf=Tf−Ts= 40C−20C= 20C
Step 3: Find the rate of temperature change at the center of the rod. The
rate of temperature change at the center of the rod is given by Fourier’s law:
dQ
dt =kA∆T
d
where: dQ
dt = Rate of heat transfer to the center of the rod k= Thermal
conductivity of the metal (given as 70 W/m
°
C) A= Cross-sectional area of the
rod = πD2
4(where Dis the diameter) ∆T= Temperature difference between
the center of the rod and the surroundings d= Length of the rod = 1.5 m
Plugging in the values: A=π·(0.02m)2
4A= 3.14 ×10−4m2
Rate of heat transfer, dQ
dt = 70 ×3.14 ×10−4×80
1.5
dQ
dt = 0.01484 W
2
Step 4: Calculate the mass of the rod. Given: Length of the rod, L= 1.5m
Diameter of the rod, D= 2 cm = 0.02 mDensity of the metal, ρ= 7,800 kg/m3
(Assuming the metal is copper)
The volume of the rod, V=πD2·L
4Mass of the rod, m=ρ·V
Calculating the mass of the rod: V=3.14×(0.02)2×1.5
4= 2.355 ×10−4m3
m= 7,800 ×2.355 ×10−4= 1.84 kg
Step 5: Calculate the specific heat energy added to the rod. Specific heat
energy, Q=mc∆T
Given: Specific heat capacity of the metal, c= 450 J/kgC
Q= 1.84 ×450 ×20 Q= 16,560 J
Step 6: Calculate the time required for the center of the rod to reach a
temperature of 40
°
C. From Step 3, we have the rate of heat transfer dQ
dt =
0.01484 W
Time, t=Q
dQ
dt
=16,560
0.01484 t= 1,114,865.05 s
Therefore, it takes approximately 1,114,865 seconds for the center of the rod
to reach a temperature of 40
°
C.
Question 3
Question
A copper cylinder of mass 0.2 kg at a temperature of 120
°
C is placed in a
calorimeter containing 0.1 kg of water at 20
°
C. If the final temperature of the
system is 25
°
C, calculate the specific heat capacity of copper. Take the specific
heat capacity of water as 4186 J/kg◦C.
Solution
Step 1: Calculate the heat lost by the copper cylinder. Given: Mass of copper
cylinder, mcopper = 0.2 kg Initial temperature of copper cylinder, Tcopper, initial =
120◦C Final temperature of the system, Tfinal = 25◦C Specific heat capacity of
water, cwater = 4186 J/kg◦C
The formula to calculate heat lost (or gained) is:
Q=mc∆T
where Qis the heat energy, mis the mass of the substance, cis the specific heat
capacity, and ∆Tis the change in temperature.
For the copper cylinder:
∆Tcopper =Tfinal −Tcopper, initial = 25◦C−120◦C = −95◦C
The heat lost by the copper cylinder is:
Qcopper lost =mcopper ·c·∆Tcopper
3
Step 2: Calculate the heat gained by the water. For the water in the
calorimeter: Initial temperature of water, Twater, initial = 20◦C
∆Twater =Tfinal −Twater, initial = 25◦C−20◦C=5◦C
The heat gained by the water is:
Qwater gained =mwater ·cwater ·∆Twater
Step 3: Set up the heat lost equals heat gained equation. Since the heat lost
by the copper is equal to the heat gained by the water (assuming no heat loss
to the surroundings), we have:
Qcopper lost =Qwater gained
Step 4: Solve for the specific heat capacity of copper, c. Substitute the ex-
pressions for Qcopper lost and Qwater gained into the equation and solve for ccopper.
mcopper ·c·∆Tcopper =mwater ·cwater ·∆Twater
Now, substitute the given values and solve for ccopper.
Question 4
Question
A metal block of mass 2 kg is initially at a temperature of 100
°
C. It is placed
in a container holding 1 kg of water at 20
°
C. Assuming no heat is lost to the
surroundings, what will be the final equilibrium temperature of the system?
(Specific heat capacity of the metal is 450 J/kg
°
C, specific heat capacity of
water is 4200 J/kg
°
C, and the latent heat of fusion for water is 334 J/g.)
Solution
Let the final equilibrium temperature of the system be T
°
C.
Step 1: Calculate heat lost by the metal block and heat gained by
the water while reaching equilibrium. The heat lost by the metal block is
equal to the heat gained by the water. Therefore,
mmetal ·cmetal ·(T−100) = mwater ·cwater ·(T−20)
Substitute the given values into the equation:
2·450 ·(T−100) = 1 ·4200 ·(T−20)
Step 2: Simplify the equation and solve for T. Expand and solve for
T:
900(T−100) = 4200(T−20)
4
900T−90000 = 4200T−84000
3300T= 60000
T=60000
3300
T≈18.18C
Therefore, the final equilibrium temperature of the system will be approxi-
mately 18.18C.
Question 5
Question
A copper block of mass 500 g and specific heat capacity c= 0.385 J/g·
°
C is
initially at a temperature of 100
°
C. It is placed in 200 g of water at 20
°
C in an
insulated container. Assuming no heat is lost to the surroundings, what will be
the final temperature of the system when the two reach thermal equilibrium?
The specific heat capacity of water is 4.18 J/g·
°
C.
Solution
Step 1: Calculate the heat lost by the copper block and the heat gained by the
water. The formula for heat transfer is given by:
Q=mc∆T
where Qis the heat transferred, mis the mass of the substance, cis the specific
heat capacity of the substance, and ∆Tis the change in temperature.
For the copper block:
Qcopper =mc∆T= (0.5 kg)(0.385 J/g ·
°
C)(Tf−100)
For the water:
Qwater =mc∆T= (0.2 kg)(4.18 J/g ·
°
C)(Tf−20)
Since the total heat gained by the water is equal to the heat lost by the
copper:
Qcopper =Qwater
Step 2: Set up and solve the equation for the final temperature (Tf).
(0.5)(0.385)(Tf−100) = (0.2)(4.18)(Tf−20)
Solving for Tf:
0.1925Tf−19.25 = 0.836Tf−16.72
0.643Tf= 2.53
Tf=2.53
0.643 ≈3.94
°
C
5
Final Answer
The final temperature of the system when the copper block and water reach
thermal equilibrium will be approximately 3.94
°
C.
Question 6
Question
A copper sphere of radius 10 cm is heated until its temperature is increased by
50
°
C. If the coefficient of linear expansion of copper is 1.7×10−5
°
C−1, calculate
the change in volume of the sphere. Assume that the bulk modulus of copper
is 140 ×109Pa.
Solution
Step 1: First, we calculate the change in volume due to the change in tempera-
ture. Given the initial temperature increase of 50
°
C, and the coefficient of linear
expansion, α= 1.7×10−5
°
C−1, the change in length can be calculated using
the formula:
∆L=α·L0·∆T
where: ∆L= change in length, α= coefficient of linear expansion, L0= initial
length, ∆T= change in temperature.
Given R= 10 cm, we have L0= 2R= 20 cm.
Substitute the values to find the change in length:
∆L= 1.7×10−5·20 cm ·50
°
C
∆L= 0.017 cm
Step 2: Next, we calculate the initial volume of the sphere. The volume of
a sphere is given by the formula:
V=4
3πr3
Given R= 10 cm, we have r=R
2= 5 cm. Substitute the value to find the
initial volume:
V0=4
3π·(5 cm)3
V0=500
3πcm3
Step 3: Now, we calculate the initial bulk modulus of the sphere. The bulk
modulus, K, is related to the Young’s modulus, Y, as:
K=Y
3(1 −2µ)
6
where: Y= 140×109Pa (given), µ= Poisson’s ratio for copper (assume µ= 0.34
for copper).
Substitute the values to find the bulk modulus:
K=140 ×109
3(1 −2×0.34)
K≈132.38 ×109Pa
Step 4: Finally, we calculate the change in volume using the formula:
∆V
V0
= 3α∆T−K
Y
Substitute the known values to find the change in volume:
∆V=V03α∆T−K
Y
∆V=500
3πcm33×1.7×10−5×50 −132.38 ×109
140 ×109
∆V≈500
3πcm3×0.00255
∆V≈2.63 cm3
Question 7
Question
A copper sphere with a radius of 5 cm is heated from an initial temperature
of 20
°
C to a final temperature of 90
°
C. If the specific heat capacity of copper
is 390 J/kg ·K and the density of copper is 8.96 g/cm3, calculate the total heat
energy required to heat the sphere.
Solution
Step 1: Determine the mass of the copper sphere. Given the density of copper
(ρCu = 8.96 g/cm3) and the radius of the sphere (r= 5 cm), we can calculate
the mass of the sphere using the formula for the volume of a sphere V=4
3πr3
and the formula for density ρ=m
V.
Volume of sphere = 4
3π(5 cm)3
=4
3π×125 cm3
=500
3πcm3
7
Thus, the mass of the copper sphere is:
m=ρCu ×V
= 8.96 ×500
3πg
Step 2: Calculate the heat energy required to heat the sphere. The heat
energy required to change the temperature of an object can be calculated using
the formula:
Q=mc∆T
where Qis the heat energy, mis the mass, cis the specific heat capacity, and
∆Tis the change in temperature.
Substitute the given values:
Q=m×390 ×(90 −20)
Step 3: Perform the final calculation. Substitute the mass calculated in Step
1 into the formula and solve for the total heat energy required.
Q= 8.96 ×500
3π×390 ×70 J
Therefore, the total heat energy required to heat the copper sphere is:
Q= 8.96 ×500
3π×390 ×70 J
Question 8
Question
A copper bar of length 2 m is initially at a uniform temperature of 100 ◦C.
One end of the bar is then immersed in steam at 100 ◦C, while the other end
is placed in an ice-water bath at 0 ◦C. The thermal conductivity of copper is
400 W/(m ·K). If the bar reaches thermal equilibrium after some time, what
will be the temperature at a distance of 1 m from the end in steam?
Solution
Step 1: We can first determine the rate at which heat is transferred through the
bar. This can be done using Fourier’s law of heat conduction:
Q=k·A·∆T
d
where: Q= rate of heat transfer, k= thermal conductivity of copper (400 W/(m ·K)),
A= cross-sectional area of the bar, ∆T= temperature difference along the bar,
and d= length of the bar.
8
Step 2: The temperature difference ∆Tcan be calculated as the difference
between the initial temperatures at the two ends of the bar:
∆T= (100 ◦C) −(0 ◦C) = 100 ◦C
Step 3: The rate at which heat is transferred through the bar can be calcu-
lated as:
Q= (400 W/(m ·K)) ·A·100 ◦C
2 m
Q= 200AW
Step 4: Since the bar eventually reaches thermal equilibrium, the same
amount of heat will be transferred at all points along the bar. This implies
that the rate of heat transfer per unit length is constant.
Step 5: If we consider a small section of length dx at a distance xfrom the
end in steam, the rate of heat transfer through this section will be constant and
equal to Q.
Step 6: The temperature Tat a distance xfrom the end in steam can be
calculated using Fourier’s law and the rate of heat transfer per unit length:
T= 100 ◦C−Q
kA ·x
Step 7: Substituting the expression for Q:
T= 100 ◦C−200A
400 W/(m ·K) ·A·x
T= 100 ◦C−1
2x
Step 8: Therefore, the temperature at a distance of 1 m from the end in
steam will be:
T= 100 ◦C−1
2(1 m) = 99.5◦C
So, the temperature at a distance of 1 m from the end in steam will be
99.5◦C.
Question 9
Question
An aluminum rod of length 0.5 m and diameter 0.02 m is initially at a tempera-
ture of 20
°
C. If 5,000 J of heat is added to the rod, calculate the final temperature
of the rod. Assume the specific heat capacity of aluminum is 900 J/kg ·K and
its density is 2700 kg/m3.
9
Solution
Step 1: Calculate the mass of the aluminum rod. The mass of the rod can be
calculated using the formula:
Volume = πr2h
where ris the radius of the rod and his the height. Given that the rod’s length
h= 0.5 m and diameter 2r= 0.02 m, we have r= 0.01 m. Therefore, the volume
of the rod is:
Volume = π(0.01 m)2×0.5 m
Volume = 5 ×10−5m3
Using the density formula Density = Mass
Volume and given the density of alu-
minum is 2700 kg/m3, the mass of the aluminum rod is:
Mass = Density ×Volume
Mass = 2700 kg/m3×5×10−5m3
Mass = 0.135 kg
Step 2: Calculate the change in temperature of the aluminum rod. The heat
added to the rod can be defined as:
Q=mc∆T
where mis the mass, cis the specific heat capacity, and ∆Tis the change in
temperature. Rearranging the formula, we have:
∆T=Q
mc
Plugging in the values Q= 5000 J, m= 0.135 kg, and c= 900 J/kg ·K, we get:
∆T=5000
0.135 ×900
∆T=5000
121.5
∆T≈41.15 K
Step 3: Calculate the final temperature of the aluminum rod. The final
temperature Tfis the initial temperature Tiplus the change in temperature
∆T. Thus:
Tf=Ti+ ∆T
Tf= 20 + 41.15
Tf≈61.15
°
C
10
Question 10
Question
A copper kettle with a mass of 2 kg contains 1 kg of water at a temperature
of 20◦C. The kettle is placed on a gas stove where the flame supplies a power
of 500 W to the kettle. The specific heat capacity of copper is 385 J/kg·K and
the specific heat capacity of water is 4186 J/kg·K. Assume no heat is lost to the
surroundings. How long will it take for the temperature of the water to reach
100◦C?
Solution
Step 1: Determine the total heat energy supplied to the system. The power
supplied to the kettle is given as 500 W. We can calculate the energy supplied
per unit time using the formula:
Energy supplied per second = Power ×Time
Step 2: Calculate the change in temperature of water. The total mass of the
system (copper kettle + water) is 3 kg. Using the formula Q=mc∆T, we can
calculate the change in temperature of water.
Step 3: Calculate the time taken to reach 100◦C. Since the copper kettle
and water are in thermal equilibrium, the heat gained by the water is equal to
the heat lost by the copper kettle. We will set up and solve an equation based
on this concept to find the time taken for the water to reach 100◦C.
Question 11
Question
A steel rod of length 2 m and diameter 2 cm is initially at a temperature of
100◦C. If it is placed in an environment at 0◦C such that the rod loses heat to
the surroundings at a rate of 200 W, how long will it take for the rod to cool
to 50◦C? Assume the thermal conductivity of steel is 50 W/(m·K), its specific
heat capacity is 500 J/(kg·K), and its density is 8000 kg/m3.
Solution
Step 1: Calculate the surface area of the rod. Given that the rod is cylindrical,
the surface area is given by:
A=π·r·h
Where ris the radius and his the height of the cylinder. Since the diameter is
2 cm, the radius is r= 1 cm = 0.01 m and the height of the cylinder is 2 m.
Therefore,
A=π·0.01 ·2=0.0628 m2
11
Step 2: Calculate the volume of the rod. The volume of the rod is given by:
V=π·r2·h
Substitute r= 0.01 m and h= 2 m to find:
V=π·0.012·2 = 0.0001 m3
Step 3: Calculate the mass of the rod. The mass of the rod can be found
using the density formula:
density = mass
volume
Rearranging for mass gives:
mass = density ·volume = 8000 ·0.0001 = 0.8 kg
Step 4: Calculate the rate of temperature change. The rate of temperature
change can be found using the formula:
P=k·A·∆T
Where Pis the power, kis the thermal conductivity, Ais the surface area, and
∆Tis the temperature change. We are given P= 200 W and k= 50 W/(m·K).
Substituting A= 0.0628 m2and rearranging for ∆T, we get:
∆T=P
k·A=200
50 ·0.0628 = 63.29 K/s
Step 5: Calculate the time taken to cool to 50◦C. To cool from 100◦C to
50◦C, the temperature change is ∆T= 100 −50 = 50 K. The time taken is
given by:
t=∆T
rate of change =50
63.29 ≈0.79 s
Question 12
Question
A copper bar of mass 0.5 kg is heated from 20
°
C to 80
°
C. If the specific heat
capacity of copper is 390 J/kg◦C, calculate the heat energy required to raise the
temperature of the copper bar.
Solution
Step 1: Determine the change in temperature Given: Initial temperature, T1=
20◦C Final temperature, T2= 80◦C
The change in temperature, ∆T=T2−T1= 80◦C−20◦C = 60◦C
12
Step 2: Calculate the heat energy using the formula The heat energy required
to raise the temperature of a substance is given by:
Q=mc∆T
where: Q= heat energy (in joules) m= mass of the substance (in kg) c= specific
heat capacity of the substance (in J/kg◦C) ∆T= change in temperature (in ◦C)
Substitute the given values into the formula:
Q= (0.5 kg) ×(390 J/kg◦C) ×(60◦C)
Step 3: Calculate the heat energy
Q= 0.5×390 ×60 = 11700 J
Therefore, the heat energy required to raise the temperature of the copper
bar is 11700 J.
Question 13
Question
A 2 kg block of copper is heated from 20
°
C to 120
°
C. Calculate the amount of
heat transferred to the block. (Specific heat capacity of copper = 390 J/kg◦C)
Solution
Step 1: Calculate the change in temperature of the copper block. Step 2: Use
the formula Q=mc∆Tto calculate the amount of heat transferred.
Step 1: The change in temperature (∆T) is given by:
∆T=Tf−Ti= 120◦C−20◦C = 100◦C
Step 2: Plug in the values into the formula Q=mc∆T:
Q= (2 kg)(390 J/kg◦C)(100◦C)
Q= 2 ×390 ×100 J
Q= 78,000 J
Therefore, the amount of heat transferred to the copper block is 78,000 J.
Question 14
Question
A 2 kg block of copper is heated until its temperature rises by 80
°
C. If the
specific heat capacity of copper is 385 J/kg
°
C, how much heat is added to the
block during this process?
13
Solution
Step 1: The formula for calculating the heat added to an object is given by:
Q=mc∆T
where: Q= heat added (in joules), m= mass of the object (in kg), c= specific
heat capacity of the material (in J/kg
°
C), ∆T= change in temperature (in
°
C).
Step 2: Substituting the given values into the formula:
Q= (2 kg)(385 J/kg
°
C)(80
°
C)
Step 3: Calculate the heat added:
Q= 2 ×385 ×80 = 61600 J
Step 4: Therefore, the heat added to the block of copper during this process
is 61,600 J.
Question 15
Question
A copper sphere with radius 10 cm is heated from 20
°
C to 100
°
C. If the coefficient
of linear expansion of copper is 1.7×10−5per
°
C and the specific heat capacity
of copper is 0.39 J/g
°
C, calculate the change in the volume of the sphere and
the heat absorbed by the sphere during this process.
Solution
Step 1: Calculate the change in volume of the sphere due to the change in
temperature. Given that the coefficient of linear expansion of copper is 1.7×10−5
per
°
C, the change in the radius of the sphere can be calculated using the
formula:
∆L=αL0∆T
where ∆Lis the change in length, αis the coefficient of linear expansion, L0is
the original length, and ∆Tis the change in temperature.
The change in volume of the sphere can be calculated using the formula:
∆V=4
3π[(R+ ∆L)3−R3]
where Ris the original radius.
Step 2: Calculate the change in volume. Given that the initial radius R=
10 cm and the change in temperature is ∆T= 100−20 = 80
°
C, we can calculate
the change in volume.
First, calculate the change in length:
∆L= 1.7×10−5×10 ×80 = 0.0136 cm
14
Then, calculate the change in volume:
∆V=4
3π[(10 + 0.0136)3−103]≈4.35 cm3
So, the change in volume of the sphere is approximately 4.35 cm3.
Step 3: Calculate the heat absorbed by the sphere during this process. The
heat absorbed by the sphere can be calculated using the formula:
Q=mc∆T
where mis the mass of the sphere and cis the specific heat capacity of copper.
The mass of the sphere can be calculated using the formula:
m=4
3πR3ρ
where ρis the density of copper.
Step 4: Calculate the heat absorbed. Given that the specific heat capacity
of copper is 0.39 J/g
°
C and the density of copper is 8.96 g/cm3, we can calculate
the mass of the sphere and then the heat absorbed.
First, calculate the mass of the sphere:
m=4
3π×103×8.96 ≈3765 g
Then, calculate the heat absorbed:
Q= 3765 ×0.39 ×80 = 117096 J
Therefore, the heat absorbed by the sphere during this process is 117096 J.
Question 16
Question
An insulated container holds 2 kg of ice at −10◦C. How much heat is required
to melt all the ice and raise the temperature of the resulting water to 20◦C?
(Specific heat capacity of ice = 2100 J/kg◦C, specific heat capacity of water
= 4186 J/kg◦C, latent heat of fusion of ice = 334 kJ/kg)
Solution
Step 1: Calculate the heat required to melt all the ice at −10◦C to water at
0◦C. The heat required to melt ice at −10◦C is given by
Q1=m·Lf= 2 kg ×334 ×103J/kg = 668000 J
15
Step 2: Calculate the heat required to raise the temperature of 2 kg of water
from 0◦C to 20◦C. The heat required is given by
Q2=m·c·∆T= 2 kg ×4186 J/kg◦C×(20 −0)◦C=2×4186 ×20 = 167440 J
Step 3: Calculate the total heat required. The total heat required is the sum
of Q1and Q2:
Qtotal =Q1+Q2= 668000 + 167440 = 835440 J
Therefore, the total heat required to melt all the ice and raise the tempera-
ture of the resulting water to 20◦C is 835440 J.
Question 17
Question
A copper block of mass 0.5 kg is initially at a temperature of 100◦C. It is
dropped into a container holding 2 kg of water at 20◦C. Assuming no heat
is lost to the surroundings, calculate the final equilibrium temperature of the
system. (Specific heat capacity of copper = 390 J/kg◦C, specific heat capacity
of water = 4186 J/kg◦C)
Solution
Step 1: Calculate the heat lost by copper to reach the final equilibrium temper-
ature.
The heat lost by an object can be calculated using the formula:
Q=mc∆T
where: - Qis the heat lost or gained, - mis the mass of the object, - cis the
specific heat capacity of the object, - ∆Tis the change in temperature.
For the copper block:
Qcopper =mcopperccopper∆Tcopper
Qcopper = 0.5 kg ×390 J/kg◦C×(Tf−100)
Step 2: Calculate the heat gained by water to reach the final equilibrium
temperature.
For the water:
Qwater =mwatercwater∆Twater
Qwater = 2 kg ×4186 J/kg◦C×(Tf−20)
Step 3: At equilibrium, the heat lost by the copper block is equal to the heat
gained by the water.
Thus, we have the equation:
0.5×390 ×(Tf−100) = 2 ×4186 ×(Tf−20)
Solve for Tfto find the final equilibrium temperature of the system.
16
Question 18
Question
A copper rod of length 1 meter and cross-sectional area 10−4m2is heated from
20◦C to 100◦C. If the thermal conductivity of copper is 400 W/mK, calculate
the heat energy transferred through the rod.
Solution
Step 1: Calculate the temperature difference Given: Initial temperature (Ti) =
20◦C Final temperature (Tf) = 100◦C
The temperature difference (∆T) is given by:
∆T=Tf−Ti= 100◦C−20◦C= 80◦C
Step 2: Calculate the rate of heat transfer The rate of heat transfer through
a material is given by Fourier’s law:
Rate of heat transfer = k·A·∆T
L
where: k= thermal conductivity of the material (400 W/mK for copper) A
= cross-sectional area of the rod (10−4m2)L= length of the rod (1 m) ∆T=
temperature difference (80◦C = 80 K)
Substitute the given values into the formula: Rate of heat transfer = 400×10−4×80
1=
32 W
Step 3: Calculate the total heat energy transferred To find the total heat
energy transferred from one end to the other end of the rod, we need to multiply
the rate of heat transfer by the time the heat is being transferred for. Since the
time is not provided in this question, we cannot calculate the total heat energy
transferred without that information.
Question 19
Question
A 0.5 kg block of aluminum at an initial temperature of 80
°
C is submerged in
2 kg of water at an initial temperature of 20
°
C. Assuming no heat is lost to the
surroundings, determine the final equilibrium temperature of the system.
Given: Specific heat capacity of aluminum, cAl = 900 J/kg ·
°
C
Specific heat capacity of water, cH2O = 4186 J/kg ·
°
C
Latent heat of fusion of ice, L= 334 J/g
Density of water, ρ= 1000 kg/m3
17
Solution
Step 1: Calculate the heat transferred from the aluminum block to the water.
The heat lost by the aluminum block is equal to the heat gained by the water,
assuming no heat is lost to the surroundings. The heat transferred is given by
the equation:
Q=mc∆T,
where mis the mass, cis the specific heat capacity, and ∆Tis the change in
temperature.
The change in temperature for the aluminum block is:
∆TAl =Tfinal −Tinitial =Tfinal −80.
The change in temperature for the water is:
∆TH2O =Tfinal −Tinitial =Tfinal −20.
Let Tfinal be the final equilibrium temperature of the system.
The heat transferred from the aluminum block to the water is:
Q=mAlcAl∆TAl =mH2OcH2O∆TH2O.
Substitute the mass and specific heat capacity values:
0.5×900 ×(Tfinal −80) = 2 ×4186 ×(Tfinal −20).
Question 20
Question
A copper block of mass 500 g at a temperature of 100
°
C is placed in a container
of water at 20
°
C. If the final equilibrium temperature of the block and water is
30
°
C, calculate the mass of water in the container. Assume specific heat capacity
of copper is 386 J/kg ·K and specific heat capacity of water is 4186 J/kg ·K.
Solution
Step 1: First, let’s calculate the heat lost by the copper block and the heat
gained by the water to reach the equilibrium temperature of 30
°
C.
The heat lost by the copper block can be calculated using the formula:
Qlost =mCu ·cCu ·∆TCu
where: - mCu is the mass of the copper block (500 g = 0.5 kg), - cCu is the
specific heat capacity of copper (386 J/kg ·K), and - ∆TCu is the change in
temperature of the copper block (100C−30C= 70K).
18
Calculating Qlost:
Qlost = 0.5 kg ·386 J/kg ·K·70K
Qlost = 13,510 J
Step 2: Next, let’s calculate the heat gained by the water using the formula:
Qgain =mH2O ·cH2O ·∆TH2O
where: - mH2O is the mass of the water in the container, - cH2O is the specific
heat capacity of water (4186 J/kg·K), and - ∆TH2O is the change in temperature
of the water (30C−20C= 10K).
Since the copper block loses heat and the water gains heat, the two values
must be equal:
Qlost =Qgain
13,510 J = mH2O ·4186 J/kg ·K·10K
Step 3: Now, solve for mH2O:
mH2O =13,510 J
41860 J/kg ·K
mH2O = 0.3227 kg
Therefore, the mass of water in the container is 322.7 g.
Question 21
Question
A 300 g block of copper at 100
°
C is dropped into 400 g of water at 20
°
C in a 200
g aluminum calorimeter cup. The initial temperature of the aluminum cup is
25
°
C. If the final temperature of the system is 30
°
C, determine the specific heat
capacity of the copper block. (Specific heat capacities: Copper = 0.387 J/g
°
C,
Water = 4.18 J/g
°
C, Aluminum = 0.897 J/g
°
C)
Solution
Step 1: Calculate the heat lost by the copper block as it cools down from 100
°
C
to the final temperature of 30
°
C. The heat lost can be calculated using the
formula:
Qcopper =mcopper ×ccopper ×∆Tcopper
where: - mcopper = 300 g is the mass of the copper block, - ccopper =
0.387 J/g
°
C is the specific heat capacity of copper, - ∆Tcopper = 100C−30C=
70Cis the change in temperature.
19
Substitute these values into the formula:
Qcopper = 300 g ×0.387 J/g
°
C×70C
Qcopper = 8139 J
Step 2: Calculate the heat gained by the aluminum cup and water as they
heat up from 25
°
C to the final temperature of 30
°
C. The heat gained by the
aluminum cup can be calculated using the formula:
QAluminum =mAluminum ×cAluminum ×∆TAluminum
where: - mAluminum = 200 g is the mass of the aluminum calorimeter cup, -
cAluminum = 0.897 J/g
°
C is the specific heat capacity of aluminum, - ∆TAluminum =
30C−25C= 5Cis the change in temperature.
Substitute these values into the formula:
QAluminum = 200 g ×0.897 J/g
°
C×5C
QAluminum = 897 J
The heat gained by the water can be calculated in a similar manner:
QWater =mWater ×cWater ×∆TWater
where: - mWater = 400 g is the mass of the water, - cWater = 4.18 J/g
°
C is
the specific heat capacity of water, - ∆TWater = 30C−25C= 5Cis the change
in temperature.
Substitute these values into the formula:
QWater = 400 g ×4.18 J/g
°
C×5C
QWater = 2090 J
Step 3: Set up the energy balance equation:
Qcopper =−(QAluminum +QWater)
Substitute the calculated values:
8139 = −(897 + 2090)
Solve for the unknown term:
8139 = −2987
This implies that the assumption we made to calculate the specific heat ca-
pacity of copper is incorrect. It is not possible to reach this condition according
to the laws of thermodynamics.
20
Question 22
Question
A 1 kg block of aluminum at an initial temperature of 20
°
C is dropped into a 1
L container of water at an initial temperature of 5
°
C. If the final temperature
of the system is 10
°
C, calculate the specific heat capacity of aluminum. Assume
the specific heat capacity of water is 4186 J/kg
°
C.
Solution
Step 1: Calculate the heat gained by water and the heat lost by aluminum
during the process.
The heat gained by water is given by the formula:
Qwater =mc∆T
where: - mis the mass of water in kg, - cis the specific heat capacity of water
in J/kg
°
C, - ∆Tis the change in temperature of water.
Given: - m= 1 kg, - c= 4186 J/kg
°
C, - ∆T= (10 −5)
°
C,
we have:
Qwater = 1 ×4186 ×(10 −5)
Qwater = 20930 J
The heat lost by aluminum is given by the formula:
Qaluminum =mc∆T
where: - mis the mass of aluminum in kg, - cis the specific heat capacity of
aluminum in J/kg
°
C, - ∆Tis the change in temperature of aluminum.
Let caluminum be the specific heat capacity of aluminum. Given: - m= 1
kg, - caluminum is the specific heat capacity of aluminum in J/kg
°
C, - ∆T=
(20 −10)
°
C,
we have:
Qaluminum = 1 ×caluminum ×(20 −10)
Qaluminum = 10caluminum J
Step 2: Apply the principle of conservation of energy.
According to the principle of conservation of energy, the heat lost by alu-
minum is equal to the heat gained by water. Therefore, we have:
Qaluminum =Qwater
10caluminum = 20930
21
caluminum =20930
10
caluminum = 2093 J/kg
°
C
Therefore, the specific heat capacity of aluminum is 2093 J/kg
°
C.
Question 23
Question
A copper kettle contains 2 kg of water at 20
°
C. How much heat is required
to raise the temperature of the water to its boiling point? The specific heat
capacity of copper is 390 J/kg◦C and the specific latent heat of vaporization of
water is 2.26 ×106J/kg.
Solution
Step 1: Calculate the heat required to raise the temperature of the water to
its boiling point. Given: - Mass of water, m= 2 kg - Initial temperature,
Ti= 20◦C - Final temperature (boiling point), Tf= 100◦C - Specific heat
capacity of water, cwater = 4186 J/kg◦C
The heat required to raise the temperature of the water can be calculated
using the formula:
Q=mcwater∆T
where ∆T=Tf−Ti.
Thus,
∆T= 100◦C−20◦C = 80◦C
Q= 2 kg ×4186 J/kg◦C×80◦C
Q= 669760 J
Step 2: Calculate the heat required for the phase change (vaporization) of
the water. The heat required for the phase change can be calculated using the
formula:
Q=mL
where Lis the specific latent heat of vaporization of water.
Given: - Specific latent heat of vaporization of water, L= 2.26 ×106J/kg
Thus,
Q= 2 kg ×2.26 ×106J/kg = 4.52 ×106J
Therefore, the total heat required is:
Total heat = Heat for temperature rise+Heat for vaporization = 669760 J+4.52×106J=5.18976×106J
22
Question 24
Question
A copper calorimeter of mass 150 g contains 200 g of water at 20
°
C. A piece of
iron of mass 100 g at 100
°
C is dropped into the calorimeter, resulting in a final
temperature of 23
°
C. Assume all the heat lost by the iron piece is gained by the
water and the calorimeter. Calculate the specific heat capacity of the iron.
Solution
Step 1: Calculate the heat lost by the iron: The heat lost by the iron can be
calculated using the formula:
Qiron =mc∆T
where mis the mass of the iron, cis the specific heat capacity of the iron, and
∆Tis the change in temperature. Substitute the given values:
Qiron = 100 g ×c×(100 −23)◦C
Qiron = 7700ccal
Step 2: Calculate the heat gained by the water and calorimeter: Since the
sum of the heat gained by the water and the calorimeter is equal to the heat
lost by the iron, we have:
Qwater +Qcalorimeter =Qiron
mwcw∆T+mccc∆T=Qiron
where mwis the mass of the water, cwis the specific heat capacity of water,
mcis the mass of the calorimeter, and ccis the specific heat capacity of the
calorimeter. Substitute the given values:
200 g ×1 cal/g◦C×(23 −20)◦C + 150 g ×cc×(23 −20)◦C = 7700c
60 + 3cc= 7700c
Step 3: Substitute the specific heat capacity of water (cw= 1 cal/g
°
C) and
solve for cc:
60 + 3cc= 7700c
3cc= 7700c−60
3cc=c(7700) −60
3cc= 7700c−60
3cc−7700c=−60
−7697cc=−60
cc=60
7697 ≈0.0078 cal/g
°
C
Therefore, the specific heat capacity of iron is approximately 0.0078 cal/g
°
C.
23
Question 25
Question
A copper sphere with a radius of 5 cm at a temperature of 200
°
C is placed in
a large cold water bath at 10
°
C. Assuming no heat is lost to the surroundings,
calculate the final temperature of the sphere after reaching thermal equilibrium.
The specific heat capacity of copper is c= 0.386 J/g
°
C, and the density of copper
is 8.96 g/cm3.
Solution
Step 1: Find the mass of the copper sphere.
The volume of a sphere is given by V=4
3πr3, where ris the radius. The density
of copper can be used to find the mass of the sphere:
mass = density ×volume
mass = 8.96 g/cm3×4
3π(5 cm)3
mass ≈1179.35 g
Step 2: Calculate the heat lost by the sphere.
The heat lost by the sphere can be calculated using the formula:
Q=mc∆T
where mis the mass, cis the specific heat capacity, and ∆Tis the change in
temperature.
Q= 1179.35 g ×0.386 J/g
°
C×(200
°
C−Tf)
Step 3: Calculate the heat gained by the sphere when reaching thermal
equilibrium.
The heat gained by the sphere can also be calculated using the formula:
Q=mc∆T
Taking the final temperature as Tfand the initial temperature as 10
°
C:
Q= 1179.35 g ×0.386 J/g
°
C×(Tf−10
°
C)
Step 4: Set the heat lost equal to the heat gained.
1179.35 g ×0.386 J/g
°
C×(200
°
C−Tf) = 1179.35 g ×0.386 J/g
°
C×(Tf−10
°
C)
Step 5: Solve for the final temperature.
141.367(200 −Tf) = 141.367(Tf−10)
24
28273.4−141.367Tf= 141.367Tf−1413.67
2(141.367Tf) = 29686.07
Tf= 105.0
°
C
Therefore, the final temperature of the copper sphere after reaching thermal
equilibrium with the water bath is 105.0
°
C.
Question 26
Question
A copper ball of mass 200 g at a temperature of 100
°
C is placed in a calorimeter
containing 400 g of water at 20
°
C. Assuming no heat is lost to the surroundings,
calculate the final temperature of the system. Specific heat capacity of copper
is 0.385 J/g◦C and specific heat capacity of water is 4.18 J/g◦C.
Solution
Step 1: Calculate the heat lost by the copper ball and the heat gained by the
water. The heat lost by the copper ball is equal to the heat gained by the water,
since no heat is lost to the surroundings. Let Tfbe the final temperature of the
system.
Heat lost by copper ball:
Qcopper = mass ×specific heat capacity ×∆T
Qcopper = 0.2 kg ×0.385 J/g◦C×(100◦C−Tf)
Heat gained by water:
Qwater = mass ×specific heat capacity ×∆T
Qwater = 0.4 kg ×4.18 J/g◦C×(Tf−20◦C)
Since no heat is lost to the surroundings, we have:
Qcopper =Qwater
Step 2: Set up and solve the equation. Set the two heat values equal and
solve for Tf:
0.2×0.385 ×(100 −Tf)=0.4×4.18 ×(Tf−20)
Solve for Tfto find the final temperature of the system.
25
Question 27
Question
In a classroom experiment, a metal cylinder initially at 100
°
C is placed in a
beaker containing 500 g of water at 20
°
C. If the final equilibrium temperature
of the system is 25
°
C, calculate the specific heat capacity of the metal cylinder,
assuming no heat is lost to the surroundings. The specific heat capacity of water
is 4.18 J/(g
°
C).
Solution
Step 1: Calculate the heat lost by the metal cylinder and the heat gained by
the water.
The heat lost by the metal cylinder is equal to the heat gained by the water.
We can use the formula:
Qlost =Qgained
where
Qlost =mc∆T
Qgained =mc∆T
Step 2: Calculate the heat lost by the metal cylinder.
Given: Initial temperature of metal cylinder, Tinitial = 100
°
C Final equilib-
rium temperature, Tfinal = 25
°
C Mass of metal cylinder, m=? (to be calculated)
Specific heat capacity of metal, c=? (to be calculated)
We can write:
Qlost =mc∆T=mc(Tfinal −Tinitial)
Step 3: Calculate the heat gained by the water.
Given: Initial temperature of water, Tinitial = 20
°
C Final equilibrium tem-
perature, Tfinal = 25
°
C Mass of water, m= 500 g Specific heat capacity of
water, c= 4.18 J/(g
°
C)
We can write:
Qgained =mc∆T=mc(Tfinal −Tinitial)
Step 4: Set up the equation and solve for the specific heat capacity of the
metal.
Since Qlost =Qgained, we have
mc(Tfinal −Tinitial) = mc(Tfinal −Tinitial)
Substitute the given values and solve for c, the specific heat capacity of the
metal cylinder.
26
Question 28
Question
A copper block of mass 0.5 kg is heated to a temperature of 100
°
C. It is then
placed in 1 kg of water at 20
°
C in an insulated container. If the final temperature
of the system is 25
°
C, calculate the specific heat capacity of copper.
Solution
Step 1: Calculate the heat lost by the copper block as it cools down from 100
°
C
to 25
°
C.
The heat lost by the copper block can be calculated using the formula:
Q=mc∆T
where: - Qis the heat lost or gained, - mis the mass of the object (0.5 kg), -
cis the specific heat capacity of copper (to be determined), - ∆Tis the change
in temperature which is 100C−25C= 75C.
Substitute the values into the formula:
Q= 0.5×c×75
Step 2: Calculate the heat gained by the water as it warms up from 20
°
C to
25
°
C.
The heat gained by the water can be calculated using the formula:
Q=mc∆T
where: - Qis the heat lost or gained, - mis the mass of the water (1 kg),
-cis the specific heat capacity of water (4186 J/kg
°
C), - ∆Tis the change in
temperature which is 25C−20C= 5C.
Substitute the values into the formula:
Q= 1 ×4186 ×5
Step 3: Since the system is insulated, the heat lost by the copper block is
equal to the heat gained by the water. Thus, equate the two heat quantities
and solve for the specific heat capacity of copper.
0.5c×75 = 1 ×4186 ×5
37.5c= 20930
c=20930
37.5
c≈557.07 J/kg
°
C
Therefore, the specific heat capacity of copper is approximately 557.07 J/kg
°
C.
27
Question 29
Question
A copper block with a mass of 0.5 kg and an initial temperature of 100
°
C is
placed in a calorimeter containing 0.2 kg of water at 20C. If the final tempera-
ture of the system is 30C, what is the specific heat capacity of the copper block?
The specific heat capacity of water is 4200 J/kg◦C.
Solution
Step 1: Calculate the heat gained by the water. The heat gained by the water
can be calculated using the formula
Qwater =mwater ·c·∆T
where: - mwater = 0.2 kg (mass of water) - c= 4200 J/kg◦C (specific heat
capacity of water) - ∆T=Tfinal −Tinitial = 30 −20 = 10
°
C Plugging in these
values, we get
Qwater = 0.2 kg ×4200 J/kg◦C×10 ◦C
Step 2: Calculate the heat lost by the copper block. The heat lost by the
copper can be calculated using the formula
Qcopper =−mcopper ·ccopper ·∆T
where: - mcopper = 0.5 kg (mass of copper) - ccopper is the specific heat capacity
of copper (to be determined) - ∆T=Tfinal −Tinitial = 30 −100 = −70
°
C
(negative because the copper is losing heat) Plugging in these values, we get
Qcopper =−0.5 kg ×ccopper × −70 ◦C
Step 3: Equate the heat gained by water to the heat lost by copper. Since
energy is conserved in this system, we have:
Qwater =−Qcopper
0.2×4200 ×10 = 0.5×ccopper ×70
Step 4: Solve for the specific heat capacity of copper. Solving the equation
gives us
ccopper =0.2×4200 ×10
0.5×70
Therefore, the specific heat capacity of the copper block is calculated to be
ccopper = 600 J/kg◦C
28
Question 30
Question
A copper block of mass 0.5 kg at a temperature of 200◦C is placed in a container
of water at 25◦C. If the final temperature of the system is 50◦C, calculate the
mass of water in the container. Assume specific heat capacity of copper is
390 J/kg◦C and specific heat capacity of water is 4186 J/kg◦C.
Solution
Step 1: Calculate the heat lost by the copper block.
The heat lost by the copper block can be calculated using the formula:
Q=mc∆T
where Qis the heat lost, mis the mass, cis the specific heat capacity, and ∆T
is the change in temperature.
Given: m= 0.5 kg c= 390 J/kg◦C Initial temperature of copper block
= 200◦C Final temperature of copper block = 50◦C
Calculating the heat lost by the copper block:
Q= 0.5×390 ×(200 −50)
Q= 0.5×390 ×150 = 29,250 J
Step 2: Calculate the heat gained by the water.
The heat gained by the water can also be calculated using the formula:
Q=mc∆T
Let Mbe the mass of water in the container. Since the water absorbs the
same amount of heat lost by the copper block,
Q=Mcw∆T
where cw= 4186 J/kg◦C is the specific heat capacity of water.
Given: cw= 4186 J/kg◦C Initial temperature of water = 25◦C Final tem-
perature of water = 50◦C
Substitute the known values and the heat lost by the copper block into the
equation:
29,250 = M×4186 ×(50 −25)
29,250 = M×4186 ×25
M=29250
104650
M≈0.279 kg
Therefore, the mass of water in the container is approximately 0.279 kg.
29
Question 31
Question
A copper block with a mass of 500 g is heated from 20
°
C to 100
°
C. If the
specific heat capacity of copper is 0.385 J/g◦C, calculate the amount of heat
energy required.
Solution
Step 1: Calculate the change in temperature. Given: Initial temperature, Ti=
20◦C Final temperature, Tf= 100◦C Change in temperature:
∆T=Tf−Ti= 100◦C−20◦C = 80◦C
Step 2: Calculate the heat energy using the formula Q=mc∆T. Given:
Mass, m= 500 g Specific heat capacity, c= 0.385 J/g◦C Change in temperature,
∆T= 80
°
C
Q=mc∆T= 500 g ×0.385 J/g◦C×80
°
C
Step 3: Calculate the heat energy.
Q= 500 ×0.385 ×80 = 15,400 J
Therefore, the amount of heat energy required to heat the copper block from
20
°
C to 100
°
C is 15,400 J.
Question 32
Question
A 500 g block of copper at 200
°
C is dropped into a container of 1 kg of water
at 20
°
C. Assuming no heat is lost to the surroundings, what will be the final
temperature of the mixture? (Specific heat capacity of copper = 0.386 J/g
°
C,
specific heat capacity of water = 4.18 J/g
°
C)
Solution
Step 1: Calculate the heat lost by the copper block. The formula for heat
transfer is given by:
Q=mc∆T
where Qis the heat transferred, mis the mass, cis the specific heat capacity,
and ∆Tis the change in temperature.
Substitute the values for the copper block into the formula:
Qcopper = 500 g ×0.386 J/g
°
C×(200 −Tf)
°
C
Qcopper = 193.0×(200 −Tf) J
30
Step 2: Calculate the heat gained by the water. Using the same formula,
but for the water:
Qwater = 1000 g ×4.18 J/g
°
C×(Tf−20)
°
C
Qwater = 4180 ×(Tf−20) J
Step 3: Set up the heat lost equals heat gained equation:
Qcopper =Qwater
193.0×(200 −Tf) = 4180 ×(Tf−20)
Step 4: Solve for the final temperature Tf.
193.0×200 −193.0×Tf= 4180 ×Tf−4180 ×20
38600 −193.0×Tf= 4180 ×Tf−83600
38600 + 83600 = 193.0×Tf+ 4180 ×Tf
122000 = 4373 ×Tf
Tf=122000
4373
Tf≈27.9C
Therefore, the final temperature of the mixture will be approximately 27.9
°
C.
Question 33
Question
A piece of metal with a mass of 500 g is heated from 20
°
C to 100
°
C. If the
specific heat capacity of the metal is 0.5 J/g
°
C, calculate the amount of heat
absorbed by the metal during this process.
Solution
Step 1: Calculate the change in temperature of the metal. Given: Initial tem-
perature, Ti= 20CFinal temperature, Tf= 100CChange in temperature,
∆T=Tf−Ti= 100C−20C= 80C
Step 2: Calculate the amount of heat absorbed. The formula for calculating
the amount of heat absorbed is:
Q=m·c·∆T
where: Q= heat absorbed (in joules) m= mass of the metal (in grams) c=
specific heat capacity of the metal (in J/g
°
C) ∆T= change in temperature of
the metal (in
°
C)
31
Substitute the given values into the formula:
Q= 500g·0.5 J/g
°
C·80C
Q= 250 J/g
°
C·80C
Q= 20000 J
Therefore, the amount of heat absorbed by the metal during this process is
20000 J.
Question 34
Question
A copper bar initially at a temperature of 100
°
C is placed in a bath of boiling
water at 100
°
C. The bar eventually reaches thermal equilibrium with the water
at 100
°
C. If the mass of the copper bar is 500 grams and the specific heat
capacity of copper is 0.385 J/g
°
C, how much heat energy was absorbed by the
copper bar? (Assume no heat is lost to the surroundings.)
Solution
Step 1: Calculate the heat energy absorbed by the copper bar using the formula:
Q=mc∆T
where: - Qis the heat energy absorbed, - mis the mass of the copper bar, - c
is the specific heat capacity of copper, and - ∆Tis the change in temperature
of the copper bar.
Step 2: Calculate the change in temperature of the copper bar using the
equation:
∆T=Tf−Ti
where: - Tfis the final temperature of the copper bar, - Tiis the initial tem-
perature of the copper bar.
Given that the initial temperature of the copper bar, Ti, is 100
°
C and the
final temperature of the copper bar, Tf, is also 100
°
C, we have:
∆T= 100C−100C= 0C
Step 3: Substitute the known values into the formula for heat energy ab-
sorbed:
Q= (500 g)(0.385 J/g
°
C)(0C)
Step 4: Calculate the heat energy absorbed by the copper bar:
Q= 0 J
Therefore, no heat energy was absorbed by the copper bar as it reached
thermal equilibrium with the boiling water at 100
°
C.
32
Question 35
Question
A copper block initially at a temperature of 100◦C is submerged in a large
container of water at 20◦C. The mass of the block is 2 kg and the specific heat
capacity of copper is 390 J/kgK. Assuming no heat is lost to the surroundings,
what is the final equilibrium temperature of the system?
Solution
Step 1: We can start by calculating the heat lost by the copper block, which
is equal to the heat gained by the water. Let’s denote the final equilibrium
temperature as Tf. The heat lost by the copper block can be calculated using
the formula:
Qcopper =mc∆T
where: - mis the mass of the copper block (2 kg), - cis the specific heat capacity
of copper (390 J/kgK), - ∆Tis the temperature change of the copper block.
Step 2: The temperature change of the copper block can be calculated as:
∆Tcopper =Tf−100
Substitute the given values to find Qcopper.
Step 3: Since the heat lost by the copper block is equal to the heat gained
by the water, we can set up an equation:
Qcopper =Qwater
The heat gained by the water can be calculated using the formula:
Qwater =mc∆T
where: - mis the mass of the water (unknown), - cis the specific heat capacity
of water (4186 J/kgK), - ∆Tis the temperature change of the water (Tf−20).
Step 4: Now, set up the equation Qcopper =Qwater and solve for Tfto find
the final equilibrium temperature of the system.
33
The heat gained by the water can be calculated as:
Qwater = 1000 ×4.18 ×(Tf−20)
Step 3: Set up the equation and solve for the final equilibrium temperature,
Tf.
Since the system is isolated and there is no heat loss to the surroundings:
Qcopper =Qwater
So, we have:
500 ×0.385 ×(100 −Tf) = 1000 ×4.18 ×(Tf−20)
Solve this equation to find the final equilibrium temperature, Tf.
Question 2
Question
A metal rod of length 1.5 m and diameter 2 cm is initially at a temperature of
100
°
C. It is then placed in a large pool of water at 20
°
C. If the heat capacity of
the metal is 450 J/kg
°
C and the thermal conductivity is 70 W/m
°
C, calculate
the time it takes for the center of the rod to reach a temperature of 40
°
C. Assume
the surroundings remain at a constant temperature of 20
°
C.
Solution
Step 1: Find the initial temperature difference between the center of the rod
and the surroundings. Given: Initial temperature of the rod, T0= 100CTem-
perature of the surroundings, Ts= 20C
The initial temperature difference, ∆T0=T0−Ts= 100C−20C= 80C
Step 2: Find the final temperature difference between the center of the rod
and the surroundings. The final temperature of the rod, Tf= 40C
The final temperature difference, ∆Tf=Tf−Ts= 40C−20C= 20C
Step 3: Find the rate of temperature change at the center of the rod. The
rate of temperature change at the center of the rod is given by Fourier’s law:
dQ
dt =kA∆T
d
where: dQ
dt = Rate of heat transfer to the center of the rod k= Thermal
conductivity of the metal (given as 70 W/m
°
C) A= Cross-sectional area of the
rod = πD2
4(where Dis the diameter) ∆T= Temperature difference between
the center of the rod and the surroundings d= Length of the rod = 1.5 m
Plugging in the values: A=π·(0.02m)2
4A= 3.14 ×10−4m2
Rate of heat transfer, dQ
dt = 70 ×3.14 ×10−4×80
1.5
dQ
dt = 0.01484 W
2
Step 4: Calculate the mass of the rod. Given: Length of the rod, L= 1.5m
Diameter of the rod, D= 2 cm = 0.02 mDensity of the metal, ρ= 7,800 kg/m3
(Assuming the metal is copper)
The volume of the rod, V=πD2·L
4Mass of the rod, m=ρ·V
Calculating the mass of the rod: V=3.14×(0.02)2×1.5
4= 2.355 ×10−4m3
m= 7,800 ×2.355 ×10−4= 1.84 kg
Step 5: Calculate the specific heat energy added to the rod. Specific heat
energy, Q=mc∆T
Given: Specific heat capacity of the metal, c= 450 J/kgC
Q= 1.84 ×450 ×20 Q= 16,560 J
Step 6: Calculate the time required for the center of the rod to reach a
temperature of 40
°
C. From Step 3, we have the rate of heat transfer dQ
dt =
0.01484 W
Time, t=Q
dQ
dt
=16,560
0.01484 t= 1,114,865.05 s
Therefore, it takes approximately 1,114,865 seconds for the center of the rod
to reach a temperature of 40
°
C.
Question 3
Question
A copper cylinder of mass 0.2 kg at a temperature of 120
°
C is placed in a
calorimeter containing 0.1 kg of water at 20
°
C. If the final temperature of the
system is 25
°
C, calculate the specific heat capacity of copper. Take the specific
heat capacity of water as 4186 J/kg◦C.
Solution
Step 1: Calculate the heat lost by the copper cylinder. Given: Mass of copper
cylinder, mcopper = 0.2 kg Initial temperature of copper cylinder, Tcopper, initial =
120◦C Final temperature of the system, Tfinal = 25◦C Specific heat capacity of
water, cwater = 4186 J/kg◦C
The formula to calculate heat lost (or gained) is:
Q=mc∆T
where Qis the heat energy, mis the mass of the substance, cis the specific heat
capacity, and ∆Tis the change in temperature.
For the copper cylinder:
∆Tcopper =Tfinal −Tcopper, initial = 25◦C−120◦C = −95◦C
The heat lost by the copper cylinder is:
Qcopper lost =mcopper ·c·∆Tcopper
3
Step 2: Calculate the heat gained by the water. For the water in the
calorimeter: Initial temperature of water, Twater, initial = 20◦C
∆Twater =Tfinal −Twater, initial = 25◦C−20◦C=5◦C
The heat gained by the water is:
Qwater gained =mwater ·cwater ·∆Twater
Step 3: Set up the heat lost equals heat gained equation. Since the heat lost
by the copper is equal to the heat gained by the water (assuming no heat loss
to the surroundings), we have:
Qcopper lost =Qwater gained
Step 4: Solve for the specific heat capacity of copper, c. Substitute the ex-
pressions for Qcopper lost and Qwater gained into the equation and solve for ccopper.
mcopper ·c·∆Tcopper =mwater ·cwater ·∆Twater
Now, substitute the given values and solve for ccopper.
Question 4
Question
A metal block of mass 2 kg is initially at a temperature of 100
°
C. It is placed
in a container holding 1 kg of water at 20
°
C. Assuming no heat is lost to the
surroundings, what will be the final equilibrium temperature of the system?
(Specific heat capacity of the metal is 450 J/kg
°
C, specific heat capacity of
water is 4200 J/kg
°
C, and the latent heat of fusion for water is 334 J/g.)
Solution
Let the final equilibrium temperature of the system be T
°
C.
Step 1: Calculate heat lost by the metal block and heat gained by
the water while reaching equilibrium. The heat lost by the metal block is
equal to the heat gained by the water. Therefore,
mmetal ·cmetal ·(T−100) = mwater ·cwater ·(T−20)
Substitute the given values into the equation:
2·450 ·(T−100) = 1 ·4200 ·(T−20)
Step 2: Simplify the equation and solve for T. Expand and solve for
T:
900(T−100) = 4200(T−20)
4
900T−90000 = 4200T−84000
3300T= 60000
T=60000
3300
T≈18.18C
Therefore, the final equilibrium temperature of the system will be approxi-
mately 18.18C.
Question 5
Question
A copper block of mass 500 g and specific heat capacity c= 0.385 J/g·
°
C is
initially at a temperature of 100
°
C. It is placed in 200 g of water at 20
°
C in an
insulated container. Assuming no heat is lost to the surroundings, what will be
the final temperature of the system when the two reach thermal equilibrium?
The specific heat capacity of water is 4.18 J/g·
°
C.
Solution
Step 1: Calculate the heat lost by the copper block and the heat gained by the
water. The formula for heat transfer is given by:
Q=mc∆T
where Qis the heat transferred, mis the mass of the substance, cis the specific
heat capacity of the substance, and ∆Tis the change in temperature.
For the copper block:
Qcopper =mc∆T= (0.5 kg)(0.385 J/g ·
°
C)(Tf−100)
For the water:
Qwater =mc∆T= (0.2 kg)(4.18 J/g ·
°
C)(Tf−20)
Since the total heat gained by the water is equal to the heat lost by the
copper:
Qcopper =Qwater
Step 2: Set up and solve the equation for the final temperature (Tf).
(0.5)(0.385)(Tf−100) = (0.2)(4.18)(Tf−20)
Solving for Tf:
0.1925Tf−19.25 = 0.836Tf−16.72
0.643Tf= 2.53
Tf=2.53
0.643 ≈3.94
°
C
5
Final Answer
The final temperature of the system when the copper block and water reach
thermal equilibrium will be approximately 3.94
°
C.
Question 6
Question
A copper sphere of radius 10 cm is heated until its temperature is increased by
50
°
C. If the coefficient of linear expansion of copper is 1.7×10−5
°
C−1, calculate
the change in volume of the sphere. Assume that the bulk modulus of copper
is 140 ×109Pa.
Solution
Step 1: First, we calculate the change in volume due to the change in tempera-
ture. Given the initial temperature increase of 50
°
C, and the coefficient of linear
expansion, α= 1.7×10−5
°
C−1, the change in length can be calculated using
the formula:
∆L=α·L0·∆T
where: ∆L= change in length, α= coefficient of linear expansion, L0= initial
length, ∆T= change in temperature.
Given R= 10 cm, we have L0= 2R= 20 cm.
Substitute the values to find the change in length:
∆L= 1.7×10−5·20 cm ·50
°
C
∆L= 0.017 cm
Step 2: Next, we calculate the initial volume of the sphere. The volume of
a sphere is given by the formula:
V=4
3πr3
Given R= 10 cm, we have r=R
2= 5 cm. Substitute the value to find the
initial volume:
V0=4
3π·(5 cm)3
V0=500
3πcm3
Step 3: Now, we calculate the initial bulk modulus of the sphere. The bulk
modulus, K, is related to the Young’s modulus, Y, as:
K=Y
3(1 −2µ)
6
where: Y= 140×109Pa (given), µ= Poisson’s ratio for copper (assume µ= 0.34
for copper).
Substitute the values to find the bulk modulus:
K=140 ×109
3(1 −2×0.34)
K≈132.38 ×109Pa
Step 4: Finally, we calculate the change in volume using the formula:
∆V
V0
= 3α∆T−K
Y
Substitute the known values to find the change in volume:
∆V=V03α∆T−K
Y
∆V=500
3πcm33×1.7×10−5×50 −132.38 ×109
140 ×109
∆V≈500
3πcm3×0.00255
∆V≈2.63 cm3
Question 7
Question
A copper sphere with a radius of 5 cm is heated from an initial temperature
of 20
°
C to a final temperature of 90
°
C. If the specific heat capacity of copper
is 390 J/kg ·K and the density of copper is 8.96 g/cm3, calculate the total heat
energy required to heat the sphere.
Solution
Step 1: Determine the mass of the copper sphere. Given the density of copper
(ρCu = 8.96 g/cm3) and the radius of the sphere (r= 5 cm), we can calculate
the mass of the sphere using the formula for the volume of a sphere V=4
3πr3
and the formula for density ρ=m
V.
Volume of sphere = 4
3π(5 cm)3
=4
3π×125 cm3
=500
3πcm3
7
Thus, the mass of the copper sphere is:
m=ρCu ×V
= 8.96 ×500
3πg
Step 2: Calculate the heat energy required to heat the sphere. The heat
energy required to change the temperature of an object can be calculated using
the formula:
Q=mc∆T
where Qis the heat energy, mis the mass, cis the specific heat capacity, and
∆Tis the change in temperature.
Substitute the given values:
Q=m×390 ×(90 −20)
Step 3: Perform the final calculation. Substitute the mass calculated in Step
1 into the formula and solve for the total heat energy required.
Q= 8.96 ×500
3π×390 ×70 J
Therefore, the total heat energy required to heat the copper sphere is:
Q= 8.96 ×500
3π×390 ×70 J
Question 8
Question
A copper bar of length 2 m is initially at a uniform temperature of 100 ◦C.
One end of the bar is then immersed in steam at 100 ◦C, while the other end
is placed in an ice-water bath at 0 ◦C. The thermal conductivity of copper is
400 W/(m ·K). If the bar reaches thermal equilibrium after some time, what
will be the temperature at a distance of 1 m from the end in steam?
Solution
Step 1: We can first determine the rate at which heat is transferred through the
bar. This can be done using Fourier’s law of heat conduction:
Q=k·A·∆T
d
where: Q= rate of heat transfer, k= thermal conductivity of copper (400 W/(m ·K)),
A= cross-sectional area of the bar, ∆T= temperature difference along the bar,
and d= length of the bar.
8
Step 2: The temperature difference ∆Tcan be calculated as the difference
between the initial temperatures at the two ends of the bar:
∆T= (100 ◦C) −(0 ◦C) = 100 ◦C
Step 3: The rate at which heat is transferred through the bar can be calcu-
lated as:
Q= (400 W/(m ·K)) ·A·100 ◦C
2 m
Q= 200AW
Step 4: Since the bar eventually reaches thermal equilibrium, the same
amount of heat will be transferred at all points along the bar. This implies
that the rate of heat transfer per unit length is constant.
Step 5: If we consider a small section of length dx at a distance xfrom the
end in steam, the rate of heat transfer through this section will be constant and
equal to Q.
Step 6: The temperature Tat a distance xfrom the end in steam can be
calculated using Fourier’s law and the rate of heat transfer per unit length:
T= 100 ◦C−Q
kA ·x
Step 7: Substituting the expression for Q:
T= 100 ◦C−200A
400 W/(m ·K) ·A·x
T= 100 ◦C−1
2x
Step 8: Therefore, the temperature at a distance of 1 m from the end in
steam will be:
T= 100 ◦C−1
2(1 m) = 99.5◦C
So, the temperature at a distance of 1 m from the end in steam will be
99.5◦C.
Question 9
Question
An aluminum rod of length 0.5 m and diameter 0.02 m is initially at a tempera-
ture of 20
°
C. If 5,000 J of heat is added to the rod, calculate the final temperature
of the rod. Assume the specific heat capacity of aluminum is 900 J/kg ·K and
its density is 2700 kg/m3.
9
Solution
Step 1: Calculate the mass of the aluminum rod. The mass of the rod can be
calculated using the formula:
Volume = πr2h
where ris the radius of the rod and his the height. Given that the rod’s length
h= 0.5 m and diameter 2r= 0.02 m, we have r= 0.01 m. Therefore, the volume
of the rod is:
Volume = π(0.01 m)2×0.5 m
Volume = 5 ×10−5m3
Using the density formula Density = Mass
Volume and given the density of alu-
minum is 2700 kg/m3, the mass of the aluminum rod is:
Mass = Density ×Volume
Mass = 2700 kg/m3×5×10−5m3
Mass = 0.135 kg
Step 2: Calculate the change in temperature of the aluminum rod. The heat
added to the rod can be defined as:
Q=mc∆T
where mis the mass, cis the specific heat capacity, and ∆Tis the change in
temperature. Rearranging the formula, we have:
∆T=Q
mc
Plugging in the values Q= 5000 J, m= 0.135 kg, and c= 900 J/kg ·K, we get:
∆T=5000
0.135 ×900
∆T=5000
121.5
∆T≈41.15 K
Step 3: Calculate the final temperature of the aluminum rod. The final
temperature Tfis the initial temperature Tiplus the change in temperature
∆T. Thus:
Tf=Ti+ ∆T
Tf= 20 + 41.15
Tf≈61.15
°
C
10
Question 10
Question
A copper kettle with a mass of 2 kg contains 1 kg of water at a temperature
of 20◦C. The kettle is placed on a gas stove where the flame supplies a power
of 500 W to the kettle. The specific heat capacity of copper is 385 J/kg·K and
the specific heat capacity of water is 4186 J/kg·K. Assume no heat is lost to the
surroundings. How long will it take for the temperature of the water to reach
100◦C?
Solution
Step 1: Determine the total heat energy supplied to the system. The power
supplied to the kettle is given as 500 W. We can calculate the energy supplied
per unit time using the formula:
Energy supplied per second = Power ×Time
Step 2: Calculate the change in temperature of water. The total mass of the
system (copper kettle + water) is 3 kg. Using the formula Q=mc∆T, we can
calculate the change in temperature of water.
Step 3: Calculate the time taken to reach 100◦C. Since the copper kettle
and water are in thermal equilibrium, the heat gained by the water is equal to
the heat lost by the copper kettle. We will set up and solve an equation based
on this concept to find the time taken for the water to reach 100◦C.
Question 11
Question
A steel rod of length 2 m and diameter 2 cm is initially at a temperature of
100◦C. If it is placed in an environment at 0◦C such that the rod loses heat to
the surroundings at a rate of 200 W, how long will it take for the rod to cool
to 50◦C? Assume the thermal conductivity of steel is 50 W/(m·K), its specific
heat capacity is 500 J/(kg·K), and its density is 8000 kg/m3.
Solution
Step 1: Calculate the surface area of the rod. Given that the rod is cylindrical,
the surface area is given by:
A=π·r·h
Where ris the radius and his the height of the cylinder. Since the diameter is
2 cm, the radius is r= 1 cm = 0.01 m and the height of the cylinder is 2 m.
Therefore,
A=π·0.01 ·2=0.0628 m2
11
Step 2: Calculate the volume of the rod. The volume of the rod is given by:
V=π·r2·h
Substitute r= 0.01 m and h= 2 m to find:
V=π·0.012·2 = 0.0001 m3
Step 3: Calculate the mass of the rod. The mass of the rod can be found
using the density formula:
density = mass
volume
Rearranging for mass gives:
mass = density ·volume = 8000 ·0.0001 = 0.8 kg
Step 4: Calculate the rate of temperature change. The rate of temperature
change can be found using the formula:
P=k·A·∆T
Where Pis the power, kis the thermal conductivity, Ais the surface area, and
∆Tis the temperature change. We are given P= 200 W and k= 50 W/(m·K).
Substituting A= 0.0628 m2and rearranging for ∆T, we get:
∆T=P
k·A=200
50 ·0.0628 = 63.29 K/s
Step 5: Calculate the time taken to cool to 50◦C. To cool from 100◦C to
50◦C, the temperature change is ∆T= 100 −50 = 50 K. The time taken is
given by:
t=∆T
rate of change =50
63.29 ≈0.79 s
Question 12
Question
A copper bar of mass 0.5 kg is heated from 20
°
C to 80
°
C. If the specific heat
capacity of copper is 390 J/kg◦C, calculate the heat energy required to raise the
temperature of the copper bar.
Solution
Step 1: Determine the change in temperature Given: Initial temperature, T1=
20◦C Final temperature, T2= 80◦C
The change in temperature, ∆T=T2−T1= 80◦C−20◦C = 60◦C
12
Step 2: Calculate the heat energy using the formula The heat energy required
to raise the temperature of a substance is given by:
Q=mc∆T
where: Q= heat energy (in joules) m= mass of the substance (in kg) c= specific
heat capacity of the substance (in J/kg◦C) ∆T= change in temperature (in ◦C)
Substitute the given values into the formula:
Q= (0.5 kg) ×(390 J/kg◦C) ×(60◦C)
Step 3: Calculate the heat energy
Q= 0.5×390 ×60 = 11700 J
Therefore, the heat energy required to raise the temperature of the copper
bar is 11700 J.
Question 13
Question
A 2 kg block of copper is heated from 20
°
C to 120
°
C. Calculate the amount of
heat transferred to the block. (Specific heat capacity of copper = 390 J/kg◦C)
Solution
Step 1: Calculate the change in temperature of the copper block. Step 2: Use
the formula Q=mc∆Tto calculate the amount of heat transferred.
Step 1: The change in temperature (∆T) is given by:
∆T=Tf−Ti= 120◦C−20◦C = 100◦C
Step 2: Plug in the values into the formula Q=mc∆T:
Q= (2 kg)(390 J/kg◦C)(100◦C)
Q= 2 ×390 ×100 J
Q= 78,000 J
Therefore, the amount of heat transferred to the copper block is 78,000 J.
Question 14
Question
A 2 kg block of copper is heated until its temperature rises by 80
°
C. If the
specific heat capacity of copper is 385 J/kg
°
C, how much heat is added to the
block during this process?
13
Solution
Step 1: The formula for calculating the heat added to an object is given by:
Q=mc∆T
where: Q= heat added (in joules), m= mass of the object (in kg), c= specific
heat capacity of the material (in J/kg
°
C), ∆T= change in temperature (in
°
C).
Step 2: Substituting the given values into the formula:
Q= (2 kg)(385 J/kg
°
C)(80
°
C)
Step 3: Calculate the heat added:
Q= 2 ×385 ×80 = 61600 J
Step 4: Therefore, the heat added to the block of copper during this process
is 61,600 J.
Question 15
Question
A copper sphere with radius 10 cm is heated from 20
°
C to 100
°
C. If the coefficient
of linear expansion of copper is 1.7×10−5per
°
C and the specific heat capacity
of copper is 0.39 J/g
°
C, calculate the change in the volume of the sphere and
the heat absorbed by the sphere during this process.
Solution
Step 1: Calculate the change in volume of the sphere due to the change in
temperature. Given that the coefficient of linear expansion of copper is 1.7×10−5
per
°
C, the change in the radius of the sphere can be calculated using the
formula:
∆L=αL0∆T
where ∆Lis the change in length, αis the coefficient of linear expansion, L0is
the original length, and ∆Tis the change in temperature.
The change in volume of the sphere can be calculated using the formula:
∆V=4
3π[(R+ ∆L)3−R3]
where Ris the original radius.
Step 2: Calculate the change in volume. Given that the initial radius R=
10 cm and the change in temperature is ∆T= 100−20 = 80
°
C, we can calculate
the change in volume.
First, calculate the change in length:
∆L= 1.7×10−5×10 ×80 = 0.0136 cm
14
Then, calculate the change in volume:
∆V=4
3π[(10 + 0.0136)3−103]≈4.35 cm3
So, the change in volume of the sphere is approximately 4.35 cm3.
Step 3: Calculate the heat absorbed by the sphere during this process. The
heat absorbed by the sphere can be calculated using the formula:
Q=mc∆T
where mis the mass of the sphere and cis the specific heat capacity of copper.
The mass of the sphere can be calculated using the formula:
m=4
3πR3ρ
where ρis the density of copper.
Step 4: Calculate the heat absorbed. Given that the specific heat capacity
of copper is 0.39 J/g
°
C and the density of copper is 8.96 g/cm3, we can calculate
the mass of the sphere and then the heat absorbed.
First, calculate the mass of the sphere:
m=4
3π×103×8.96 ≈3765 g
Then, calculate the heat absorbed:
Q= 3765 ×0.39 ×80 = 117096 J
Therefore, the heat absorbed by the sphere during this process is 117096 J.
Question 16
Question
An insulated container holds 2 kg of ice at −10◦C. How much heat is required
to melt all the ice and raise the temperature of the resulting water to 20◦C?
(Specific heat capacity of ice = 2100 J/kg◦C, specific heat capacity of water
= 4186 J/kg◦C, latent heat of fusion of ice = 334 kJ/kg)
Solution
Step 1: Calculate the heat required to melt all the ice at −10◦C to water at
0◦C. The heat required to melt ice at −10◦C is given by
Q1=m·Lf= 2 kg ×334 ×103J/kg = 668000 J
15
Step 2: Calculate the heat required to raise the temperature of 2 kg of water
from 0◦C to 20◦C. The heat required is given by
Q2=m·c·∆T= 2 kg ×4186 J/kg◦C×(20 −0)◦C=2×4186 ×20 = 167440 J
Step 3: Calculate the total heat required. The total heat required is the sum
of Q1and Q2:
Qtotal =Q1+Q2= 668000 + 167440 = 835440 J
Therefore, the total heat required to melt all the ice and raise the tempera-
ture of the resulting water to 20◦C is 835440 J.
Question 17
Question
A copper block of mass 0.5 kg is initially at a temperature of 100◦C. It is
dropped into a container holding 2 kg of water at 20◦C. Assuming no heat
is lost to the surroundings, calculate the final equilibrium temperature of the
system. (Specific heat capacity of copper = 390 J/kg◦C, specific heat capacity
of water = 4186 J/kg◦C)
Solution
Step 1: Calculate the heat lost by copper to reach the final equilibrium temper-
ature.
The heat lost by an object can be calculated using the formula:
Q=mc∆T
where: - Qis the heat lost or gained, - mis the mass of the object, - cis the
specific heat capacity of the object, - ∆Tis the change in temperature.
For the copper block:
Qcopper =mcopperccopper∆Tcopper
Qcopper = 0.5 kg ×390 J/kg◦C×(Tf−100)
Step 2: Calculate the heat gained by water to reach the final equilibrium
temperature.
For the water:
Qwater =mwatercwater∆Twater
Qwater = 2 kg ×4186 J/kg◦C×(Tf−20)
Step 3: At equilibrium, the heat lost by the copper block is equal to the heat
gained by the water.
Thus, we have the equation:
0.5×390 ×(Tf−100) = 2 ×4186 ×(Tf−20)
Solve for Tfto find the final equilibrium temperature of the system.
16
Question 18
Question
A copper rod of length 1 meter and cross-sectional area 10−4m2is heated from
20◦C to 100◦C. If the thermal conductivity of copper is 400 W/mK, calculate
the heat energy transferred through the rod.
Solution
Step 1: Calculate the temperature difference Given: Initial temperature (Ti) =
20◦C Final temperature (Tf) = 100◦C
The temperature difference (∆T) is given by:
∆T=Tf−Ti= 100◦C−20◦C= 80◦C
Step 2: Calculate the rate of heat transfer The rate of heat transfer through
a material is given by Fourier’s law:
Rate of heat transfer = k·A·∆T
L
where: k= thermal conductivity of the material (400 W/mK for copper) A
= cross-sectional area of the rod (10−4m2)L= length of the rod (1 m) ∆T=
temperature difference (80◦C = 80 K)
Substitute the given values into the formula: Rate of heat transfer = 400×10−4×80
1=
32 W
Step 3: Calculate the total heat energy transferred To find the total heat
energy transferred from one end to the other end of the rod, we need to multiply
the rate of heat transfer by the time the heat is being transferred for. Since the
time is not provided in this question, we cannot calculate the total heat energy
transferred without that information.
Question 19
Question
A 0.5 kg block of aluminum at an initial temperature of 80
°
C is submerged in
2 kg of water at an initial temperature of 20
°
C. Assuming no heat is lost to the
surroundings, determine the final equilibrium temperature of the system.
Given: Specific heat capacity of aluminum, cAl = 900 J/kg ·
°
C
Specific heat capacity of water, cH2O = 4186 J/kg ·
°
C
Latent heat of fusion of ice, L= 334 J/g
Density of water, ρ= 1000 kg/m3
17
Solution
Step 1: Calculate the heat transferred from the aluminum block to the water.
The heat lost by the aluminum block is equal to the heat gained by the water,
assuming no heat is lost to the surroundings. The heat transferred is given by
the equation:
Q=mc∆T,
where mis the mass, cis the specific heat capacity, and ∆Tis the change in
temperature.
The change in temperature for the aluminum block is:
∆TAl =Tfinal −Tinitial =Tfinal −80.
The change in temperature for the water is:
∆TH2O =Tfinal −Tinitial =Tfinal −20.
Let Tfinal be the final equilibrium temperature of the system.
The heat transferred from the aluminum block to the water is:
Q=mAlcAl∆TAl =mH2OcH2O∆TH2O.
Substitute the mass and specific heat capacity values:
0.5×900 ×(Tfinal −80) = 2 ×4186 ×(Tfinal −20).
Question 20
Question
A copper block of mass 500 g at a temperature of 100
°
C is placed in a container
of water at 20
°
C. If the final equilibrium temperature of the block and water is
30
°
C, calculate the mass of water in the container. Assume specific heat capacity
of copper is 386 J/kg ·K and specific heat capacity of water is 4186 J/kg ·K.
Solution
Step 1: First, let’s calculate the heat lost by the copper block and the heat
gained by the water to reach the equilibrium temperature of 30
°
C.
The heat lost by the copper block can be calculated using the formula:
Qlost =mCu ·cCu ·∆TCu
where: - mCu is the mass of the copper block (500 g = 0.5 kg), - cCu is the
specific heat capacity of copper (386 J/kg ·K), and - ∆TCu is the change in
temperature of the copper block (100C−30C= 70K).
18
Calculating Qlost:
Qlost = 0.5 kg ·386 J/kg ·K·70K
Qlost = 13,510 J
Step 2: Next, let’s calculate the heat gained by the water using the formula:
Qgain =mH2O ·cH2O ·∆TH2O
where: - mH2O is the mass of the water in the container, - cH2O is the specific
heat capacity of water (4186 J/kg·K), and - ∆TH2O is the change in temperature
of the water (30C−20C= 10K).
Since the copper block loses heat and the water gains heat, the two values
must be equal:
Qlost =Qgain
13,510 J = mH2O ·4186 J/kg ·K·10K
Step 3: Now, solve for mH2O:
mH2O =13,510 J
41860 J/kg ·K
mH2O = 0.3227 kg
Therefore, the mass of water in the container is 322.7 g.
Question 21
Question
A 300 g block of copper at 100
°
C is dropped into 400 g of water at 20
°
C in a 200
g aluminum calorimeter cup. The initial temperature of the aluminum cup is
25
°
C. If the final temperature of the system is 30
°
C, determine the specific heat
capacity of the copper block. (Specific heat capacities: Copper = 0.387 J/g
°
C,
Water = 4.18 J/g
°
C, Aluminum = 0.897 J/g
°
C)
Solution
Step 1: Calculate the heat lost by the copper block as it cools down from 100
°
C
to the final temperature of 30
°
C. The heat lost can be calculated using the
formula:
Qcopper =mcopper ×ccopper ×∆Tcopper
where: - mcopper = 300 g is the mass of the copper block, - ccopper =
0.387 J/g
°
C is the specific heat capacity of copper, - ∆Tcopper = 100C−30C=
70Cis the change in temperature.
19
Substitute these values into the formula:
Qcopper = 300 g ×0.387 J/g
°
C×70C
Qcopper = 8139 J
Step 2: Calculate the heat gained by the aluminum cup and water as they
heat up from 25
°
C to the final temperature of 30
°
C. The heat gained by the
aluminum cup can be calculated using the formula:
QAluminum =mAluminum ×cAluminum ×∆TAluminum
where: - mAluminum = 200 g is the mass of the aluminum calorimeter cup, -
cAluminum = 0.897 J/g
°
C is the specific heat capacity of aluminum, - ∆TAluminum =
30C−25C= 5Cis the change in temperature.
Substitute these values into the formula:
QAluminum = 200 g ×0.897 J/g
°
C×5C
QAluminum = 897 J
The heat gained by the water can be calculated in a similar manner:
QWater =mWater ×cWater ×∆TWater
where: - mWater = 400 g is the mass of the water, - cWater = 4.18 J/g
°
C is
the specific heat capacity of water, - ∆TWater = 30C−25C= 5Cis the change
in temperature.
Substitute these values into the formula:
QWater = 400 g ×4.18 J/g
°
C×5C
QWater = 2090 J
Step 3: Set up the energy balance equation:
Qcopper =−(QAluminum +QWater)
Substitute the calculated values:
8139 = −(897 + 2090)
Solve for the unknown term:
8139 = −2987
This implies that the assumption we made to calculate the specific heat ca-
pacity of copper is incorrect. It is not possible to reach this condition according
to the laws of thermodynamics.
20
Question 22
Question
A 1 kg block of aluminum at an initial temperature of 20
°
C is dropped into a 1
L container of water at an initial temperature of 5
°
C. If the final temperature
of the system is 10
°
C, calculate the specific heat capacity of aluminum. Assume
the specific heat capacity of water is 4186 J/kg
°
C.
Solution
Step 1: Calculate the heat gained by water and the heat lost by aluminum
during the process.
The heat gained by water is given by the formula:
Qwater =mc∆T
where: - mis the mass of water in kg, - cis the specific heat capacity of water
in J/kg
°
C, - ∆Tis the change in temperature of water.
Given: - m= 1 kg, - c= 4186 J/kg
°
C, - ∆T= (10 −5)
°
C,
we have:
Qwater = 1 ×4186 ×(10 −5)
Qwater = 20930 J
The heat lost by aluminum is given by the formula:
Qaluminum =mc∆T
where: - mis the mass of aluminum in kg, - cis the specific heat capacity of
aluminum in J/kg
°
C, - ∆Tis the change in temperature of aluminum.
Let caluminum be the specific heat capacity of aluminum. Given: - m= 1
kg, - caluminum is the specific heat capacity of aluminum in J/kg
°
C, - ∆T=
(20 −10)
°
C,
we have:
Qaluminum = 1 ×caluminum ×(20 −10)
Qaluminum = 10caluminum J
Step 2: Apply the principle of conservation of energy.
According to the principle of conservation of energy, the heat lost by alu-
minum is equal to the heat gained by water. Therefore, we have:
Qaluminum =Qwater
10caluminum = 20930
21
caluminum =20930
10
caluminum = 2093 J/kg
°
C
Therefore, the specific heat capacity of aluminum is 2093 J/kg
°
C.
Question 23
Question
A copper kettle contains 2 kg of water at 20
°
C. How much heat is required
to raise the temperature of the water to its boiling point? The specific heat
capacity of copper is 390 J/kg◦C and the specific latent heat of vaporization of
water is 2.26 ×106J/kg.
Solution
Step 1: Calculate the heat required to raise the temperature of the water to
its boiling point. Given: - Mass of water, m= 2 kg - Initial temperature,
Ti= 20◦C - Final temperature (boiling point), Tf= 100◦C - Specific heat
capacity of water, cwater = 4186 J/kg◦C
The heat required to raise the temperature of the water can be calculated
using the formula:
Q=mcwater∆T
where ∆T=Tf−Ti.
Thus,
∆T= 100◦C−20◦C = 80◦C
Q= 2 kg ×4186 J/kg◦C×80◦C
Q= 669760 J
Step 2: Calculate the heat required for the phase change (vaporization) of
the water. The heat required for the phase change can be calculated using the
formula:
Q=mL
where Lis the specific latent heat of vaporization of water.
Given: - Specific latent heat of vaporization of water, L= 2.26 ×106J/kg
Thus,
Q= 2 kg ×2.26 ×106J/kg = 4.52 ×106J
Therefore, the total heat required is:
Total heat = Heat for temperature rise+Heat for vaporization = 669760 J+4.52×106J=5.18976×106J
22
Question 24
Question
A copper calorimeter of mass 150 g contains 200 g of water at 20
°
C. A piece of
iron of mass 100 g at 100
°
C is dropped into the calorimeter, resulting in a final
temperature of 23
°
C. Assume all the heat lost by the iron piece is gained by the
water and the calorimeter. Calculate the specific heat capacity of the iron.
Solution
Step 1: Calculate the heat lost by the iron: The heat lost by the iron can be
calculated using the formula:
Qiron =mc∆T
where mis the mass of the iron, cis the specific heat capacity of the iron, and
∆Tis the change in temperature. Substitute the given values:
Qiron = 100 g ×c×(100 −23)◦C
Qiron = 7700ccal
Step 2: Calculate the heat gained by the water and calorimeter: Since the
sum of the heat gained by the water and the calorimeter is equal to the heat
lost by the iron, we have:
Qwater +Qcalorimeter =Qiron
mwcw∆T+mccc∆T=Qiron
where mwis the mass of the water, cwis the specific heat capacity of water,
mcis the mass of the calorimeter, and ccis the specific heat capacity of the
calorimeter. Substitute the given values:
200 g ×1 cal/g◦C×(23 −20)◦C + 150 g ×cc×(23 −20)◦C = 7700c
60 + 3cc= 7700c
Step 3: Substitute the specific heat capacity of water (cw= 1 cal/g
°
C) and
solve for cc:
60 + 3cc= 7700c
3cc= 7700c−60
3cc=c(7700) −60
3cc= 7700c−60
3cc−7700c=−60
−7697cc=−60
cc=60
7697 ≈0.0078 cal/g
°
C
Therefore, the specific heat capacity of iron is approximately 0.0078 cal/g
°
C.
23
Question 25
Question
A copper sphere with a radius of 5 cm at a temperature of 200
°
C is placed in
a large cold water bath at 10
°
C. Assuming no heat is lost to the surroundings,
calculate the final temperature of the sphere after reaching thermal equilibrium.
The specific heat capacity of copper is c= 0.386 J/g
°
C, and the density of copper
is 8.96 g/cm3.
Solution
Step 1: Find the mass of the copper sphere.
The volume of a sphere is given by V=4
3πr3, where ris the radius. The density
of copper can be used to find the mass of the sphere:
mass = density ×volume
mass = 8.96 g/cm3×4
3π(5 cm)3
mass ≈1179.35 g
Step 2: Calculate the heat lost by the sphere.
The heat lost by the sphere can be calculated using the formula:
Q=mc∆T
where mis the mass, cis the specific heat capacity, and ∆Tis the change in
temperature.
Q= 1179.35 g ×0.386 J/g
°
C×(200
°
C−Tf)
Step 3: Calculate the heat gained by the sphere when reaching thermal
equilibrium.
The heat gained by the sphere can also be calculated using the formula:
Q=mc∆T
Taking the final temperature as Tfand the initial temperature as 10
°
C:
Q= 1179.35 g ×0.386 J/g
°
C×(Tf−10
°
C)
Step 4: Set the heat lost equal to the heat gained.
1179.35 g ×0.386 J/g
°
C×(200
°
C−Tf) = 1179.35 g ×0.386 J/g
°
C×(Tf−10
°
C)
Step 5: Solve for the final temperature.
141.367(200 −Tf) = 141.367(Tf−10)
24
28273.4−141.367Tf= 141.367Tf−1413.67
2(141.367Tf) = 29686.07
Tf= 105.0
°
C
Therefore, the final temperature of the copper sphere after reaching thermal
equilibrium with the water bath is 105.0
°
C.
Question 26
Question
A copper ball of mass 200 g at a temperature of 100
°
C is placed in a calorimeter
containing 400 g of water at 20
°
C. Assuming no heat is lost to the surroundings,
calculate the final temperature of the system. Specific heat capacity of copper
is 0.385 J/g◦C and specific heat capacity of water is 4.18 J/g◦C.
Solution
Step 1: Calculate the heat lost by the copper ball and the heat gained by the
water. The heat lost by the copper ball is equal to the heat gained by the water,
since no heat is lost to the surroundings. Let Tfbe the final temperature of the
system.
Heat lost by copper ball:
Qcopper = mass ×specific heat capacity ×∆T
Qcopper = 0.2 kg ×0.385 J/g◦C×(100◦C−Tf)
Heat gained by water:
Qwater = mass ×specific heat capacity ×∆T
Qwater = 0.4 kg ×4.18 J/g◦C×(Tf−20◦C)
Since no heat is lost to the surroundings, we have:
Qcopper =Qwater
Step 2: Set up and solve the equation. Set the two heat values equal and
solve for Tf:
0.2×0.385 ×(100 −Tf)=0.4×4.18 ×(Tf−20)
Solve for Tfto find the final temperature of the system.
25
Question 27
Question
In a classroom experiment, a metal cylinder initially at 100
°
C is placed in a
beaker containing 500 g of water at 20
°
C. If the final equilibrium temperature
of the system is 25
°
C, calculate the specific heat capacity of the metal cylinder,
assuming no heat is lost to the surroundings. The specific heat capacity of water
is 4.18 J/(g
°
C).
Solution
Step 1: Calculate the heat lost by the metal cylinder and the heat gained by
the water.
The heat lost by the metal cylinder is equal to the heat gained by the water.
We can use the formula:
Qlost =Qgained
where
Qlost =mc∆T
Qgained =mc∆T
Step 2: Calculate the heat lost by the metal cylinder.
Given: Initial temperature of metal cylinder, Tinitial = 100
°
C Final equilib-
rium temperature, Tfinal = 25
°
C Mass of metal cylinder, m=? (to be calculated)
Specific heat capacity of metal, c=? (to be calculated)
We can write:
Qlost =mc∆T=mc(Tfinal −Tinitial)
Step 3: Calculate the heat gained by the water.
Given: Initial temperature of water, Tinitial = 20
°
C Final equilibrium tem-
perature, Tfinal = 25
°
C Mass of water, m= 500 g Specific heat capacity of
water, c= 4.18 J/(g
°
C)
We can write:
Qgained =mc∆T=mc(Tfinal −Tinitial)
Step 4: Set up the equation and solve for the specific heat capacity of the
metal.
Since Qlost =Qgained, we have
mc(Tfinal −Tinitial) = mc(Tfinal −Tinitial)
Substitute the given values and solve for c, the specific heat capacity of the
metal cylinder.
26
Question 28
Question
A copper block of mass 0.5 kg is heated to a temperature of 100
°
C. It is then
placed in 1 kg of water at 20
°
C in an insulated container. If the final temperature
of the system is 25
°
C, calculate the specific heat capacity of copper.
Solution
Step 1: Calculate the heat lost by the copper block as it cools down from 100
°
C
to 25
°
C.
The heat lost by the copper block can be calculated using the formula:
Q=mc∆T
where: - Qis the heat lost or gained, - mis the mass of the object (0.5 kg), -
cis the specific heat capacity of copper (to be determined), - ∆Tis the change
in temperature which is 100C−25C= 75C.
Substitute the values into the formula:
Q= 0.5×c×75
Step 2: Calculate the heat gained by the water as it warms up from 20
°
C to
25
°
C.
The heat gained by the water can be calculated using the formula:
Q=mc∆T
where: - Qis the heat lost or gained, - mis the mass of the water (1 kg),
-cis the specific heat capacity of water (4186 J/kg
°
C), - ∆Tis the change in
temperature which is 25C−20C= 5C.
Substitute the values into the formula:
Q= 1 ×4186 ×5
Step 3: Since the system is insulated, the heat lost by the copper block is
equal to the heat gained by the water. Thus, equate the two heat quantities
and solve for the specific heat capacity of copper.
0.5c×75 = 1 ×4186 ×5
37.5c= 20930
c=20930
37.5
c≈557.07 J/kg
°
C
Therefore, the specific heat capacity of copper is approximately 557.07 J/kg
°
C.
27
Question 29
Question
A copper block with a mass of 0.5 kg and an initial temperature of 100
°
C is
placed in a calorimeter containing 0.2 kg of water at 20C. If the final tempera-
ture of the system is 30C, what is the specific heat capacity of the copper block?
The specific heat capacity of water is 4200 J/kg◦C.
Solution
Step 1: Calculate the heat gained by the water. The heat gained by the water
can be calculated using the formula
Qwater =mwater ·c·∆T
where: - mwater = 0.2 kg (mass of water) - c= 4200 J/kg◦C (specific heat
capacity of water) - ∆T=Tfinal −Tinitial = 30 −20 = 10
°
C Plugging in these
values, we get
Qwater = 0.2 kg ×4200 J/kg◦C×10 ◦C
Step 2: Calculate the heat lost by the copper block. The heat lost by the
copper can be calculated using the formula
Qcopper =−mcopper ·ccopper ·∆T
where: - mcopper = 0.5 kg (mass of copper) - ccopper is the specific heat capacity
of copper (to be determined) - ∆T=Tfinal −Tinitial = 30 −100 = −70
°
C
(negative because the copper is losing heat) Plugging in these values, we get
Qcopper =−0.5 kg ×ccopper × −70 ◦C
Step 3: Equate the heat gained by water to the heat lost by copper. Since
energy is conserved in this system, we have:
Qwater =−Qcopper
0.2×4200 ×10 = 0.5×ccopper ×70
Step 4: Solve for the specific heat capacity of copper. Solving the equation
gives us
ccopper =0.2×4200 ×10
0.5×70
Therefore, the specific heat capacity of the copper block is calculated to be
ccopper = 600 J/kg◦C
28
Question 30
Question
A copper block of mass 0.5 kg at a temperature of 200◦C is placed in a container
of water at 25◦C. If the final temperature of the system is 50◦C, calculate the
mass of water in the container. Assume specific heat capacity of copper is
390 J/kg◦C and specific heat capacity of water is 4186 J/kg◦C.
Solution
Step 1: Calculate the heat lost by the copper block.
The heat lost by the copper block can be calculated using the formula:
Q=mc∆T
where Qis the heat lost, mis the mass, cis the specific heat capacity, and ∆T
is the change in temperature.
Given: m= 0.5 kg c= 390 J/kg◦C Initial temperature of copper block
= 200◦C Final temperature of copper block = 50◦C
Calculating the heat lost by the copper block:
Q= 0.5×390 ×(200 −50)
Q= 0.5×390 ×150 = 29,250 J
Step 2: Calculate the heat gained by the water.
The heat gained by the water can also be calculated using the formula:
Q=mc∆T
Let Mbe the mass of water in the container. Since the water absorbs the
same amount of heat lost by the copper block,
Q=Mcw∆T
where cw= 4186 J/kg◦C is the specific heat capacity of water.
Given: cw= 4186 J/kg◦C Initial temperature of water = 25◦C Final tem-
perature of water = 50◦C
Substitute the known values and the heat lost by the copper block into the
equation:
29,250 = M×4186 ×(50 −25)
29,250 = M×4186 ×25
M=29250
104650
M≈0.279 kg
Therefore, the mass of water in the container is approximately 0.279 kg.
29
Question 31
Question
A copper block with a mass of 500 g is heated from 20
°
C to 100
°
C. If the
specific heat capacity of copper is 0.385 J/g◦C, calculate the amount of heat
energy required.
Solution
Step 1: Calculate the change in temperature. Given: Initial temperature, Ti=
20◦C Final temperature, Tf= 100◦C Change in temperature:
∆T=Tf−Ti= 100◦C−20◦C = 80◦C
Step 2: Calculate the heat energy using the formula Q=mc∆T. Given:
Mass, m= 500 g Specific heat capacity, c= 0.385 J/g◦C Change in temperature,
∆T= 80
°
C
Q=mc∆T= 500 g ×0.385 J/g◦C×80
°
C
Step 3: Calculate the heat energy.
Q= 500 ×0.385 ×80 = 15,400 J
Therefore, the amount of heat energy required to heat the copper block from
20
°
C to 100
°
C is 15,400 J.
Question 32
Question
A 500 g block of copper at 200
°
C is dropped into a container of 1 kg of water
at 20
°
C. Assuming no heat is lost to the surroundings, what will be the final
temperature of the mixture? (Specific heat capacity of copper = 0.386 J/g
°
C,
specific heat capacity of water = 4.18 J/g
°
C)
Solution
Step 1: Calculate the heat lost by the copper block. The formula for heat
transfer is given by:
Q=mc∆T
where Qis the heat transferred, mis the mass, cis the specific heat capacity,
and ∆Tis the change in temperature.
Substitute the values for the copper block into the formula:
Qcopper = 500 g ×0.386 J/g
°
C×(200 −Tf)
°
C
Qcopper = 193.0×(200 −Tf) J
30
Step 2: Calculate the heat gained by the water. Using the same formula,
but for the water:
Qwater = 1000 g ×4.18 J/g
°
C×(Tf−20)
°
C
Qwater = 4180 ×(Tf−20) J
Step 3: Set up the heat lost equals heat gained equation:
Qcopper =Qwater
193.0×(200 −Tf) = 4180 ×(Tf−20)
Step 4: Solve for the final temperature Tf.
193.0×200 −193.0×Tf= 4180 ×Tf−4180 ×20
38600 −193.0×Tf= 4180 ×Tf−83600
38600 + 83600 = 193.0×Tf+ 4180 ×Tf
122000 = 4373 ×Tf
Tf=122000
4373
Tf≈27.9C
Therefore, the final temperature of the mixture will be approximately 27.9
°
C.
Question 33
Question
A piece of metal with a mass of 500 g is heated from 20
°
C to 100
°
C. If the
specific heat capacity of the metal is 0.5 J/g
°
C, calculate the amount of heat
absorbed by the metal during this process.
Solution
Step 1: Calculate the change in temperature of the metal. Given: Initial tem-
perature, Ti= 20CFinal temperature, Tf= 100CChange in temperature,
∆T=Tf−Ti= 100C−20C= 80C
Step 2: Calculate the amount of heat absorbed. The formula for calculating
the amount of heat absorbed is:
Q=m·c·∆T
where: Q= heat absorbed (in joules) m= mass of the metal (in grams) c=
specific heat capacity of the metal (in J/g
°
C) ∆T= change in temperature of
the metal (in
°
C)
31
Substitute the given values into the formula:
Q= 500g·0.5 J/g
°
C·80C
Q= 250 J/g
°
C·80C
Q= 20000 J
Therefore, the amount of heat absorbed by the metal during this process is
20000 J.
Question 34
Question
A copper bar initially at a temperature of 100
°
C is placed in a bath of boiling
water at 100
°
C. The bar eventually reaches thermal equilibrium with the water
at 100
°
C. If the mass of the copper bar is 500 grams and the specific heat
capacity of copper is 0.385 J/g
°
C, how much heat energy was absorbed by the
copper bar? (Assume no heat is lost to the surroundings.)
Solution
Step 1: Calculate the heat energy absorbed by the copper bar using the formula:
Q=mc∆T
where: - Qis the heat energy absorbed, - mis the mass of the copper bar, - c
is the specific heat capacity of copper, and - ∆Tis the change in temperature
of the copper bar.
Step 2: Calculate the change in temperature of the copper bar using the
equation:
∆T=Tf−Ti
where: - Tfis the final temperature of the copper bar, - Tiis the initial tem-
perature of the copper bar.
Given that the initial temperature of the copper bar, Ti, is 100
°
C and the
final temperature of the copper bar, Tf, is also 100
°
C, we have:
∆T= 100C−100C= 0C
Step 3: Substitute the known values into the formula for heat energy ab-
sorbed:
Q= (500 g)(0.385 J/g
°
C)(0C)
Step 4: Calculate the heat energy absorbed by the copper bar:
Q= 0 J
Therefore, no heat energy was absorbed by the copper bar as it reached
thermal equilibrium with the boiling water at 100
°
C.
32
Question 35
Question
A copper block initially at a temperature of 100◦C is submerged in a large
container of water at 20◦C. The mass of the block is 2 kg and the specific heat
capacity of copper is 390 J/kgK. Assuming no heat is lost to the surroundings,
what is the final equilibrium temperature of the system?
Solution
Step 1: We can start by calculating the heat lost by the copper block, which
is equal to the heat gained by the water. Let’s denote the final equilibrium
temperature as Tf. The heat lost by the copper block can be calculated using
the formula:
Qcopper =mc∆T
where: - mis the mass of the copper block (2 kg), - cis the specific heat capacity
of copper (390 J/kgK), - ∆Tis the temperature change of the copper block.
Step 2: The temperature change of the copper block can be calculated as:
∆Tcopper =Tf−100
Substitute the given values to find Qcopper.
Step 3: Since the heat lost by the copper block is equal to the heat gained
by the water, we can set up an equation:
Qcopper =Qwater
The heat gained by the water can be calculated using the formula:
Qwater =mc∆T
where: - mis the mass of the water (unknown), - cis the specific heat capacity
of water (4186 J/kgK), - ∆Tis the temperature change of the water (Tf−20).
Step 4: Now, set up the equation Qcopper =Qwater and solve for Tfto find
the final equilibrium temperature of the system.
33
