PHYS 232 - UNIVERSITY PHYSICS II Temperature and heat Question Bank Set 8

PHYS 232 - UNIVERSITY PHYSICS

II - Temperature and heat

Question Bank - Set 8

Liberty University

Question 1

Question

A copper rod of length 2 m and cross-sectional area 0.01 m2is initially at a

temperature of 100◦C. It is then placed in an ice bath at 0◦C. Calculate the

amount of heat transferred from the copper rod to the ice bath. (Specific heat

capacity of copper = 390 J/kg ·◦C, density of copper = 8960 kg/m3, latent heat

of fusion of ice = 334 kJ/kg).

Solution

Step 1: Calculate the mass of the copper rod. Given density of copper, ρ=

8960 kg/m3; length, L= 2 m; cross-sectional area, A= 0.01 m2. The volume

of the rod, V=A·L. The mass of the rod, m=ρ·V. Therefore, m=

8960 kg/m3×0.01 m2×2m.m= 179.2kg.

Step 2: Calculate the initial thermal energy of the copper rod. The initial

temperature of the copper rod, Ti= 100◦C = 100 + 273 = 373 K. The specific

heat capacity of copper, c= 390 J/kg◦C. The initial thermal energy, Qi=

m·c·∆T. Here, ∆T=Tf−Ti= 0◦C−100◦C = −100◦C. Thus, Qi=

179.2kg ×390 J/kg ·◦C×(−100 ◦C). Qi=−7.008 ×106J.

Step 3: Calculate the heat required to change the temperature of ice. The

latent heat of fusion of ice, Lf= 334 ×103J/kg. The heat required to change

the temperature of ice, Qf=m×Lf. Therefore, Qf= 179.2kg×334×103J/kg.

Qf= 5.9752 ×107J.

Step 4: Calculate the total amount of heat transferred. The total heat

transferred, Qtotal =Qi+Qf. Thus, Qtotal =−7.008 ×106J+ 5.9752 ×107J.

Qtotal = 5.2744 ×107J.

Therefore, the amount of heat transferred from the copper rod to the ice

bath is 5.2744 ×107J.

Question 2

Question

A copper rod of length 2.0 m and radius 0.5 cm is heated from an initial tem-

perature of 20

C to a final temperature of 80

C. If the coefficient of linear ex-

pansion of copper is 1.7×10−5per degree Celsius and its specific heat capacity

is 0.385 J/g

C, calculate the heat transferred to the rod. Assume the rod is

perfectly insulated and that the temperature is uniform throughout.

Solution

Step 1: Calculate the change in length of the copper rod due to heating. Given

the coefficient of linear expansion of copper, α= 1.7×10−5

C−1, the change in

length, ∆L, can be calculated using the formula:

∆L=α·L·∆T

where Lis the initial length, ∆Tis the change in temperature. Substituting

the values:

∆L= (1.7×10−5)×2.0 m ×(80 −20)

C=0.0068 m

Step 2: Calculate the volume expansion of the copper rod. The volume

expansion can be calculated using the formula for the volume of a cylinder:

V=πr2h

where ris the initial radius and his the initial length. The change in volume,

∆V, is given by:

∆V=π(r+ ∆r)2−r2L=πr2+ 2r∆r+ (∆r)2−r2L=π·2r∆r·L

Substituting the values:

∆V=π·2(0.005 m)(0.0068 m) ·2 m = 0.00043 m3

Step 3: Calculate the heat transferred to the rod. The heat transferred can

be calculated using the formula:

Q=mc∆T

where mis the mass of the copper rod, cis the specific heat capacity of cop-

per, and ∆Tis the change in temperature. Since the density of copper is

8,960 kg/m3, the mass of the rod can be calculated as:

m=ρV = 8960 kg/m3×0.00043 m3= 3.848 kg

Substitute the values to find the heat transferred:

Q= 3.848 kg ×0.385 J/g

C×(80 −20)

C = 355.08 J

Question 3

Question

An aluminum rod of length 1 m and cross-sectional area 0.01 m2is heated from

20◦C to 80◦C. The rod has a thermal conductivity of 200 W/(m

K), a specific

heat capacity of 900 J/(kg

K), and a density of 2700 kg/m3. Calculate the

amount of heat transferred to the rod during this process.

Solution

Step 1: Calculate the mass of the aluminum rod. Given the density of aluminum

is 2700 kg/m3and the volume of the rod is V= 1 m ×0.01 m2= 0.01 m3, the

mass mcan be calculated as:

m= density ×volume = 2700 kg/m3×0.01 m3= 27 kg

Step 2: Calculate the change in temperature. The change in temperature

∆T=Tf−Ti= 80◦C−20◦C = 60 K.

Step 3: Calculate the amount of heat transferred. The amount of heat

transferred Qis given by the formula:

Q=mc∆T

Substitute the mass m= 27 kg, specific heat capacity c= 900 J/(kg

K), and

change in temperature ∆T= 60 K:

Q= 27 kg ×900 J/(kg

K) ×60 K = 1458000 J

Therefore, the amount of heat transferred to the aluminum rod during this

process is 1,458,000 J.

Question 4

Question

A copper rod of length 1.5 m and cross-sectional area 4 cm2is heated from

C to 80

C. If the coefficient of linear expansion of copper is 1.7×10−5

C−1

and the specific heat capacity of copper is 0.387 J/g

C, calculate the change in

length of the rod and the heat absorbed by the rod.

Solution

Step 1: Calculate the change in length of the rod due to heating. Given the

initial temperature T1= 20

C, final temperature T2= 80

C, and coefficient

of linear expansion α= 1.7×10−5

C−1, the change in length is given by the

formula:

∆L=L·α·∆T

where ∆T=T2−T1and Lis the original length of the rod.

Substitute the given values:

∆L= 1.5 m ·1.7×10−5

C−1·(80 −20)

∆L= 1.5×10−2m=0.015 m

Step 2: Calculate the heat absorbed by the rod. The heat absorbed by the

rod can be calculated using the formula:

Q=mc∆T

where mis the mass of the rod, cis the specific heat capacity of copper, and

∆Tis the change in temperature.

First, calculate the mass of the rod using the formula for volume:

V=A·L

m=ρ·V

where ρ= 8.96 g/cm3is the density of copper.

Substitute the values to find the mass m:

m= 8.96 g/cm3·4 cm2·1.5 m ·104cm2/m2

m= 538.56 kg = 538560 g

Now, substitute the mass m= 538560 g, specific heat capacity c= 0.387

J/g

C, and change in temperature ∆T= 80 −20 = 60

C into the formula:

Q= 538560 g ·0.387 J/g◦C·60

Q= 12324480 J

Therefore, the change in length of the rod is 0.015 m and the heat absorbed

by the rod is 12324480 J.

Question 5

Question

A copper rod of length 2 meters and diameter 1 cm is heated from 20

C to

C. If the thermal conductivity of copper is 401 W/(m.K), calculate the heat

conducted through the rod during this process.

Solution

Let’s first calculate the surface area of the rod. The formula for the surface area

of a cylinder is given by:

A= 2πrh + 2πr2

where ris the radius of the rod and his the length of the rod.

Step 1: Calculate the radius of the rod. Given that the diameter of the rod

is 1 cm, the radius ris half of the diameter.

r=1

2×(1 cm) = 0.5 cm = 0.005 m

Step 2: Calculate the surface area of the rod.

A= 2π×0.005 ×2+2π×0.0052

A= 0.063 m2

Step 3: Calculate the temperature difference. The temperature difference

is 80 ◦C−20 ◦C = 60 ◦C = 60 K.

Step 4: Calculate the heat transfer. The formula for heat transfer is given

by:

Q=k·A·∆T

where Qis the heat transfer, kis the thermal conductivity, Ais the surface

area, and ∆Tis the temperature difference.

Substitute the values into the formula:

Q= 401 W/(m.K) ×0.063 m2×60 K

Q= 1519.4 W

Therefore, the heat conducted through the rod during this process is 1519.4

Watts.

Question 6

Question

A copper cylinder with a radius of 5 cm and a height of 10 cm is initially at a

temperature of 100

C. It is then placed in a water bath at 20

C. If the specific

heat capacity of copper is 0.386 J/g

C, how much heat is transferred to the water

bath when the copper cylinder reaches thermal equilibrium with the water?

Solution

Step 1: Calculate the mass of the copper cylinder. The volume of the cylinder

can be calculated using the formula for the volume of a cylinder: V=πr2h,

where ris the radius and his the height. Substituting r= 5 cm and h= 10

cm:

V=π×(5 cm)2×10 cm = 250πcm3

The density of copper is 8.96 g/cm

, so the mass mof the cylinder is:

m= density ×volume = 8.96 g/cm3×250πcm3

Step 2: Calculate the change in temperature of the copper cylinder. The

heat lost by the copper cylinder equals the heat gained by the water bath. The

formula for heat transfer is Q=mc∆T, where Qis the heat transferred, mis

the mass, cis the specific heat capacity, and ∆Tis the change in temperature.

The change in temperature of the copper cylinder is ∆T=Tfinal −Tinitial.

Let Tfinal be the final temperature when the copper cylinder reaches thermal

equilibrium with the water bath. Therefore, ∆T=Tfinal −100 ◦C.

Step 3: Calculate the heat transferred to the water bath. Since the heat lost

by the copper cylinder is equal to the heat gained by the water bath:

mcopper ×ccopper ×∆T=mwater ×cwater ×∆T

Substitute in the given values for the specific heat capacities of copper and

water, and solve for mwater:

mwater =mcopper ×ccopper

cwater

=8.96 ×250π×0.386

4.18

Calculate the final temperature Tfinal by rearranging the heat transfer formula:

Tfinal =mcopper ×ccopper ×100 + mwater ×cwater ×20

mcopper ×ccopper +mwater ×cwater

Finally, calculate the heat transferred to the water bath once the copper

cylinder reaches thermal equilibrium using the formula Q=mwater ×cwater ×

(Tfinal −20).

Question 7

Question

A copper rod of length 2 meters and diameter 1 cm has one face maintained at

a temperature of 100

C while the other face is maintained at 0

C. If the thermal

conductivity of copper is 390 W/mK and the density is 8900 kg/m3, calculate

the rate of heat conducted through the rod.

Solution

Let’s denote the rate of heat conducted through the rod as Q. We can use

Fourier’s law of heat conduction, which states dQ

dt =kAdT

dx , where kis the ther-

mal conductivity, Ais the cross-sectional area, dT

dx is the temperature gradient

along the rod, and dQ

dt is the rate of heat conducted through the rod.

Step 1: Find the cross-sectional area of the rod The cross-sectional

area of the rod, denoted as A, can be calculated using the formula A=πd2

where dis the diameter of the rod. Given that the diameter is 1 cm, we have

d= 0.01 m. Therefore, the cross-sectional area is:

A=π·(0.01)2

4= 7.85 ×10−5m2

Step 2: Calculate the temperature gradient The temperature gradient,

dx , is the change in temperature per unit length. In this case, it is 100−0

C/m.

Step 3: Calculate the rate of heat conducted through the rod

Plugging the values into Fourier’s law of heat conduction:

dt =kAdT

dt = 390 ×7.85 ×10−5×50

dt = 0.3073 W

Therefore, the rate of heat conducted through the rod is 0.3073 W.

Question 8

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 100◦C? (Specific heat capacity of copper

= 0.386 J/g

C, specific heat capacity of water = 4.18 J/g

C).

Solution

Step 1: First, calculate the heat required to raise the temperature of the water

from 20◦C to 100◦C using the formula:

Q=mc∆T

where mis the mass of the water, cis the specific heat capacity of water, and

∆Tis the change in temperature.

Step 2: Calculate the heat required to raise the temperature of the water:

Qwater = 2000g×4.18J/gC ×(100 −20)C

Step 3: Calculate the heat absorbed by the water:

Qwater = 2000g×4.18J/gC ×80C

Step 4: Next, calculate the heat absorbed by the kettle using the formula:

Q=mc∆T

where mis the mass of the kettle, cis the specific heat capacity of copper, and

∆Tis the change in temperature.

Step 5: Assume that the kettle and the water reach thermal equilibrium at

100◦C, so the heat absorbed by the kettle is equal to the heat lost by the water.

Step 6: Let Qkettle be the heat absorbed by the kettle. Using the same

formula as before:

Qkettle =mc∆T

Qwater =Qkettle

Step 7: Set the two equations equal to each other and solve for the mass of

the kettle:

2000g×4.18J/gC ×80C=m×0.386J/gC ×80C

Step 8: Calculate the heat required to raise the temperature of the water to

100◦C by adding the heat absorbed by the water and the heat absorbed by the

kettle:

Q=Qwater +Qkettle

Question 9

Question

A copper rod of length 1.5 m and cross-sectional area 0.02m2is initially at a

temperature of 100◦C. It is immersed in a large water bath at a constant temper-

ature of 20◦C. Given that the thermal conductivity of copper is 400W/(m·K),

determine how long it takes for the rod to cool to 50◦C. Assume one-dimensional

heat conduction and neglect heat losses to the surroundings.

Solution

Step 1: Find the rate of heat loss from the rod using Fourier’s law. The rate of

heat loss (dQ

dt ) through the rod is given by Fourier’s law:

dt =−kAdT

where kis the thermal conductivity of copper, Ais the cross-sectional area of

the rod, Tis the temperature of the rod, and xis the distance along the rod.

At T= 100◦Cand x= 0, we have:

dt =−(400)(0.02)100 −20

1.5

dt =−20 W

Step 2: Use the energy balance equation to find the time taken for the rod

to cool to 50◦C. The total heat lost by the rod when it cools from 100◦Cto

50◦Cis:

Q=mc∆T

where mis the mass of the rod, cis the specific heat capacity of copper, and

∆Tis the temperature change. The mass of the rod can be calculated using the

formula m= volume ×density:

m=ALρ = (0.02)(1.5)(8,960) = 270.4 kg

The specific heat capacity of copper is c= 390 J/(kg ·◦C) and the temperature

change is ∆T= 100 −50 = 50. Thus, the total heat lost is:

Q= (270.4)(390)(50) = 5,283,600 J

The time taken to lose this amount of heat can be found using:

Q=Zt

dt dt

5,283,600 = Zt

−20 dt

5,283,600 = −20t

t=−264,180 s

t= 73.4 hours

Therefore, it takes approximately 73.4 hours for the rod to cool from 100◦Cto

50◦C.

Question 10

Question

A 350 g piece of copper at a temperature of 90

C is placed in 500 g of water

at 20

C. Assuming no heat is lost to the surroundings, what will be the final

temperature of the system? (Specific heat capacity of copper = 0.385 J/g

specific heat capacity of water = 4.18 J/g

Solution

Step 1: Calculate the heat absorbed by the copper piece as it cools down to the

final temperature: The formula for heat transfer is Q=mc∆T, where: - mis

the mass of the substance, - cis the specific heat capacity of the substance, and

- ∆Tis the change in temperature.

Given: Initial temperature of copper (Tcopper) = 90

C, Final temperature of

the system (Tfinal) = ?.

Let the final temperature of the system be Tfinal. The change in temperature

for the copper is ∆T=Tcopper −Tfinal = 90 −Tfinal. So, the heat absorbed by

the copper piece is:

Qcopper =mc(∆T) = 350 ×0.385 ×(90 −Tfinal)

Step 2: Calculate the heat released by the water as it warms up to the final

temperature: Similar to the copper piece, the heat released by the water can be

calculated using the formula Q=mc∆T. The change in temperature for the

water is ∆T=Tfinal −20. So, the heat released by the water is:

Qwater =mc(∆T) = 500 ×4.18 ×(Tfinal −20)

Step 3: Since there is no heat lost to the surroundings, the heat lost by the

copper must be equal to the heat gained by the water. Thus, we have:

Qcopper =Qwater

350 ×0.385 ×(90 −Tfinal) = 500 ×4.18 ×(Tfinal −20)

Step 4: Solve the equation obtained in Step 3 to find the final temperature

of the system.

350 ×0.385 ×90 −350 ×0.385 ×Tfinal = 500 ×4.18 ×Tfinal −500 ×4.18 ×20

12555 −134.75Tfinal = 2090Tfinal −8360

Solving for Tfinal:

134.75Tfinal + 2090Tfinal = 12555 + 8360

2224.75Tfinal = 20915

Tfinal ≈9.4C

Therefore, the final temperature of the system will be approximately 9.4

Question 11

Question

A metal block at 100

C is placed into a container of water at 20

C. The block

has a mass of 2 kg and a specific heat capacity of 450 J/kg

C. If the final

temperature of the system is 30

C, calculate the mass of water in the container.

The specific heat capacity of water is 4186 J/kg

Solution

Step 1: Calculate the heat lost by the metal block.

Given that the initial temperature of the metal block is 100

C, the final

temperature is 30

C, and the specific heat capacity is 450 J/kg

C, we can use

the formula for heat energy:

Qmetal =mc∆T

Qmetal = 2 ×450 ×(30 −100)

Qmetal = 2 ×450 ×(−70)

Qmetal =−63000 J

Step 2: Calculate the heat gained by the water.

Given that the water is at an initial temperature of 20

C and a final tem-

perature of 30

C, and the specific heat capacity of water is 4186 J/kg

C, we can

use the formula for heat energy:

Qwater =mc∆T

Qwater =m×4186 ×(30 −20)

Qwater =m×4186 ×10

Qwater = 41860mJ

Step 3: Set up the equation based on the conservation of energy principle:

−63000 = 41860m

m=−63000

41860

m=−1.5 kg

Since mass cannot be negative, we made an error in the calculations. Let’s

correct it:

Step 4: Correct the calculation for mass of water.

m=63000

41860

m≈1.50 kg

Therefore, the mass of the water in the container is approximately 1.50 kg.

Question 12

Question

A piece of copper with mass 200 g is heated to 100◦C and then placed in a

beaker containing 500 g of water at 20◦C. The final temperature of the system

is 25◦C. Assuming no heat is lost to the surroundings, determine the specific

heat capacity of copper. The specific heat capacity of water is 4200 J/kg·K.

Solution

Step 1: Calculate the heat absorbed by the copper when heated to 100◦C.

Heat absorbed = mc∆T

= (0.2 kg)(390 J/kg ·K)(100 K)

= 780 J

Step 2: Calculate the heat lost by the copper to reach the final temperature

of 25◦C.

Heat lost = mc∆T

= (0.2 kg)c(75 K)

Step 3: Calculate the heat gained by the water to reach the final temperature

of 25◦C.

Heat gained by water = mc∆T

= (0.5 kg)(4200 J/kg ·K)(5 K)

= 10500 J

Step 4: Since the system is thermally isolated, the heat lost by the copper

is equal to the heat gained by the water.

0.2c(75) = 10500

c=10500

15 = 700 J/kg ·K

Therefore, the specific heat capacity of copper is 700 J/kg·K.

Question 13

Question

A 50 g piece of copper at 100

C is dropped into 200 g of water at 20

C in an

insulated container. Assuming no heat is lost to the surroundings, calculate the

final temperature of the water and copper. Specific heat capacity of copper is

0.385 J/g

C and specific heat capacity of water is 4.18 J/g

Solution

Step 1: Calculate the heat lost by the copper and gained by the water 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.

For the copper:

Qcopper = (50 g)(0.385 J/g

C)(Tf−100)

= 19.25Tf−1925

For the water:

Qwater = (200 g)(4.18 J/g

C)(Tf−20)

= 836Tf−16720

Since the heat lost by the copper is equal to the heat gained by the water

(assuming no heat is lost to the surroundings), we have:

19.25Tf−1925 = 836Tf−16720

Step 2: Solve for the final temperature Tf.

19.25Tf−1925 = 836Tf−16720

836Tf−19.25Tf= 16720 −1925

816.75Tf= 14795

Tf=14795

816.75

Tf≈18.10C

Therefore, the final temperature of the water and copper is approximately

18.10

Question 14

Question

A copper rod of length 2 m and cross-sectional area 0.003 m2is heated at one

end. The temperature at that end is maintained at 100◦C while the temperature

at the other end is maintained at 0◦C. If the rod dissipates heat uniformly at

a rate of 600 W, what is the thermal conductivity of copper? (Specific heat

capacity of copper = 390 J/kg·K, density of copper = 8960 kg/m3)

Solution

Step 1: Calculate the rate of heat flow through the rod. Given that the rate at

which heat is dissipated is 600 W, the rate at which heat flows through the rod

is also 600 W.

Step 2: Calculate the temperature difference (∆T) across the rod. Since one

end is at 100◦C and the other end is at 0◦C, the temperature difference across

the rod is:

∆T= 100 −0 = 100 ◦C

∆T= 100 + 273 = 373 K

Step 3: Calculate the rate of heat flow through the rod using the formula

for heat flow:

Q=k·A·∆T

where Qis the rate of heat flow, kis the thermal conductivity, Ais the cross-

sectional area, ∆Tis the temperature difference, and Lis the length of the

rod.

Step 4: Substitute the known values into the formula and solve for the

thermal conductivity (k):

600 = k·0.003 ·373

600 = 1.119k

1200 = 1.119k

k≈1200

1.119

k≈1071.63 W/(m ·K)

Therefore, the thermal conductivity of copper is approximately 1071.63

W/(m·K).

Question 15

Question

An aluminum block of mass 2 kg is initially at a temperature of 200◦C. It is

placed in a container of water at 25◦C. The aluminum block reaches thermal

equilibrium with the water after some time. Assuming no heat is lost to the

surroundings, calculate the final temperature of the system.

Given: Specific heat capacity of aluminum = 900 J/kg◦C Specific heat ca-

pacity of water = 4200 J/kg◦C Mass of water = 5 kg Initial temperature of

water = 25◦C

Solution

Step 1: Calculate the heat lost by the aluminum block and the heat gained by

the water. The total heat lost by the aluminum block = The total heat gained

by the water (Qlost =Qgain).

Let Tfbe the final temperature of the system.

To calculate the heat lost by the aluminum block: Qaluminum =maluminum ·

caluminum ·(Tf−200)

To calculate the heat gained by the water: Qwater =mwater ·cwater ·(Tf−25)

Setting Qlost =Qgain: 2 ·900 ·(Tf−200) = 5 ·4200 ·(Tf−25)

Step 2: Solve the equation to find the final temperature (Tf) of the system.

Solving the equation we obtained in step 1: 1800Tf−360000 = 21000Tf−52500

Simplifying: 21000Tf−1800Tf= 52500 −360000

19200Tf=−307500

Tf=−307500

19200 ≈ −16.016

Since temperature cannot be negative, this result is not physically meaning-

ful.

Therefore, the final temperature of the system cannot be determined without

additional information or considerations.

Question 16

Question

A copper rod of length 2 m and cross-sectional area 4 cm2is initially at a

temperature of 200◦C. If heat is supplied to the rod at a rate of 400 W, find

the steady-state temperature at the center of the rod. Assume the thermal

conductivity of copper is 390 W/(m ·K) and neglect any heat losses.

Solution

Step 1: Find the thermal resistance of the copper rod. The thermal resistance

Rof a material can be calculated using the formula:

R=L

where: - Lis the length of the rod, - kis the thermal conductivity of the material

(given as 390 W/(m ·K) for copper), - Ais the cross-sectional area of the rod.

Substitute L= 2 m, k= 390 W/(m ·K) and A= 4 ×10−4m2:

R=2

390 ×4×10−4= 12.82 K/W

Step 2: Calculate the heat equation for the rod. The rate of heat transfer

through the rod can be calculated using the formula:

P=∆Tcenter

where: - Pis the power of heat supplied to the rod (given as 400 W), - ∆Tcenter

is the temperature difference across the length of the rod, - Ris the thermal

resistance of the rod.

Substitute P= 400 W and R= 12.82 K/W:

400 = ∆Tcenter/12.82

∆Tcenter = 400 ×12.82 = 5128 K

Step 3: Find the steady-state temperature at the center of the rod. Since

the rod is initially at 200◦C, the steady-state temperature at the center of the

rod will be:

Tcenter = 200 + ∆Tcenter

2= 200 + 5128

2= 2664 K

Therefore, the steady-state temperature at the center of the copper rod is

2664 K.

Question 17

Question

A block of copper with a mass of 0.5 kg is initially at a temperature of 100◦C.

It is placed in a container with 2 kg of water at 10◦C. Assuming no heat is

lost to the surroundings, what will be the final temperature of the system when

thermal equilibrium is reached? (Specific heat capacity of copper is 390 J/kg·K

and specific heat capacity of water is 4186 J/kg·K)

Solution

Step 1: First, we need to calculate the heat lost by the copper block and the

heat gained by the water until they reach thermal equilibrium.

The heat lost by the copper block can be calculated using the formula:

Q=mc∆T

where: Q= heat lost by the copper block, m= mass of the copper block, c=

specific heat capacity of copper, ∆T= change in temperature for the copper

block.

Plugging in the values, we get:

Qcopper = 0.5 kg ×390 J/kg·K×(Tfinal −100)

Step 2: The heat gained by the water can be calculated using the formula:

Q=mc∆T

where: Q= heat gained by the water, m= mass of the water, c= specific heat

capacity of water, ∆T= change in temperature for the water.

Plugging in the values, we get:

Qwater = 2 kg ×4186 J/kg·K×(Tfinal −10)

Step 3: Since the system is isolated (no heat is lost to the surroundings),

the heat lost by the copper block is equal to the heat gained by the water:

0.5×390 ×(Tfinal −100) = 2 ×4186 ×(Tfinal −10)

Step 4: Solve the equation for Tfinal:

195(Tfinal −100) = 8372(Tfinal −10)

195Tfinal −19500 = 8372Tfinal −83720

8177Tfinal = 64220

Tfinal =64220

8177 ≈7.85◦C

Therefore, the final temperature of the system when thermal equilibrium is

reached is approximately 7.85◦C.

Question 18

Question

A block of metal with mass 2 kg and specific heat capacity 0.5 J/g

C is initially

at a temperature of 100

C. It is submerged in 1 kg of water at 20

C. If the

final temperature of the system is 30

C, what is the specific heat capacity of the

metal?

Solution

Step 1: Calculate the heat lost by the metal.

The heat lost by the metal is given by the formula:

Qmetal =mc∆T

where m= 2 kg (mass of the metal), cis the specific heat capacity of the metal,

∆T= (100 −30) = 70

C (initial temperature - final temperature).

Substitute in the values to find Qmetal:

Qmetal = 2 kg ×c×70

Step 2: Calculate the heat gained by the water.

The heat gained by the water is given by the formula:

Qwater =mc∆T

where m= 1 kg (mass of the water), c= 4.18 J/g

C (specific heat capacity of

water), ∆T= (30 −20) = 10

C (final temperature - initial temperature).

Substitute in the values to find Qwater:

Qwater = 1 kg ×4.18 J/g

C×10

Step 3: Since the heat lost by the metal is equal to the heat gained by the

water (assuming no heat loss to the surroundings), we can set up the equation:

Qmetal =Qwater

Step 4: Equate the expressions for Qmetal and Qwater, and solve for the

specific heat capacity of the metal:

2 kg ×c×70

C = 1 kg ×4.18 J/g

C×10

Step 5: Solve for c, the specific heat capacity of the metal:

c=1 kg ×4.18 J/g

C×10

2 kg ×70

Question 19

Question

A copper rod of length 2 m and cross-sectional area 0.01 m2is heated from 20

to 70

C. If the density of copper is 8900 kg/m3and its specific heat capacity is

386 J/kg

C, calculate the heat energy supplied to the rod.

Solution

Let’s break down the problem into several steps:

Step 1: Find the mass of the copper rod using the formula:

mass = density ×volume

Step 2: Calculate the change in temperature of the copper rod:

∆T=Tfinal −Tinitial

Step 3: Determine the heat energy supplied to the rod using the formula:

Q=mc∆T

Step 1: Find the mass of the copper rod: Given:

Density of copper (ρ) = 8900 kg/m3

Cross-sectional area (A) = 0.01 m2

Length (L) = 2 m

The volume of the rod is given by

Volume (V) = area ×length = AL

Thus,

mass = ρ×Volume = ρAL

Substitute the given values to find the mass of the copper rod.

Step 2: Calculate the change in temperature: Given:

Tinitial = 20◦C

Tfinal = 70◦C

Calculate the change in temperature:

∆T=Tfinal −Tinitial

Step 3: Determine the heat energy supplied to the rod: Given:

Specific heat capacity (c) = 386 J/kg◦C

Use the formula for heat energy:

Q=mc∆T

Substitute the calculated mass, specific heat capacity, and change in tem-

perature to find the heat energy supplied to the rod.

Question 20

Question

A copper pot contains 2 kg of water at 20

C. How much heat must be added

to the water to bring it to boiling point and convert it to steam at 100

C? The

specific heat capacity of water is 4186 J/kg

C, the specific latent heat of fusion

of water is 334 J/g, and the latent heat of vaporization of water is 2260 J/g.

Solution

Let’s break down the process into several steps:



Heating the water from 20

C to 100



Changing the water at 100

C to steam at 100

Step 1: Heating the water from 20

C to 100

The formula to calculate the heat energy required to change the temperature

of a substance is:

Q=mc∆T

where: Q= heat energy (Joules) m= mass of the substance (kg) c= specific

heat capacity of the substance (J/kg

C) ∆T= change in temperature (

Given that the initial temperature T1= 20Cand the final temperature

T2= 100C, we can calculate ∆T:

∆T=T2−T1= 100C−20C= 80C

Now, we can calculate the heat energy Q1required to heat the water:

Q1= (2 kg)(4186 J/kg

C)(80C)

Q1= 669760 J

Step 2: Changing the water at 100

C to steam at 100

To convert water at 100

C to steam at 100

C, we need to account for the

latent heat of vaporization:

Q2=mL

where: Q2= heat energy (Joules) m= mass of water (kg) L= latent heat

of vaporization (J/kg)

Given that the specific latent heat of fusion of water is 334 J/g, we need to

convert this to J/kg:

L= 334 J/g ×1000 g/kg = 334000 J/kg

Now we can calculate the heat energy Q2required to convert the water to

steam:

Q2= (2 kg)(334000 J/kg)

Q2= 668000 J

Finally, the total heat energy Qtotal required is the sum of Q1and Q2:

Qtotal =Q1+Q2= 669760 J + 668000 J = 1339760 J

Thus, 1.34 ×106J of heat must be added to the water to bring it to boiling

point and convert it to steam at 100

Question 21

Question

A steel rod of length 2.5 m and diameter 2 cm is initially at a uniform temper-

ature of 100◦C. If 5.0 x 105J of heat is supplied to the rod, calculate the final

temperature of the rod. Assume the specific heat capacity of steel is 450J/kg◦C

and the density of steel is 7800 kg/m3.

Solution

Step 1: Firstly, we need to calculate the mass of the steel rod using the formula

m=π×d2×L×ρ

4, where dis the diameter, Lis the length, and ρis the density

of steel. Given d= 2 cm = 0.02 m, L= 2.5 m, and ρ= 7800 kg/m3, we have:

m=π×(0.02 m)2×2.5 m ×7800 kg/m3

m≈0.6124 kg

Step 2: Next, we can calculate the change in temperature using the formula

Q=mc∆T, where Qis the heat supplied, mis the mass, cis the specific

heat capacity, and ∆Tis the change in temperature. Given Q= 5.0×105J,

m= 0.6124 kg, and c= 450 J/kg◦C, we rearrange the formula to solve for ∆T:

∆T=Q

mc =5.0×105J

0.6124 kg ×450 J/kg◦C

∆T≈2222.2◦C

Step 3: Finally, we can calculate the final temperature by adding the initial

temperature to the change in temperature. Given initial temperature = 100◦C,

we have:

Tfinal = 100 + 2222.2

Tfinal ≈2322.2◦C

Therefore, the final temperature of the steel rod after the heat transfer is

approximately 2322.2◦C.

Question 22

Question

A copper rod of length 2.0 m and diameter 1.0 cm is initially at a temperature of

C. How much heat needs to be added to increase its temperature to 100

Assume the specific heat capacity of copper is 0.385 J/g

C and its density is

8.96 g/cm3.

Solution

Step 1: First, calculate the mass of the copper rod. Given that the diameter of

the rod is 1.0 cm, the radius ris 0.5 cm. We can convert the radius to meters

by dividing by 100: r= 0.5 cm = 0.005 m.

The volume Vof the rod can be calculated using the formula for the volume

of a cylinder: V=πr2h, where h= 2.0 m is the height of the rod.

Substitute the values to find the volume: V=π×(0.005 m)2×2.0 m ⇒V=

π×0.000025 m2×2.0 m ⇒V= 0.000157 m3.

The mass mof the rod can be calculated using the density formula m=

density ×volume: m= 8.96 g/cm3×0.000157 m3⇒m= 1.40752 kg.

Step 2: Calculate the heat required to increase the temperature of the copper

rod. The heat Qrequired can be calculated using the formula: Q=mc∆T,

where mis the mass of the rod, cis the specific heat capacity of copper, and

∆Tis the change in temperature.

Given: m= 1.40752 kg, c= 0.385 J/g

C = 385 J/kg

C, ∆T= 100C−20C=

80C.

Substitute the values into the formula: Q= 1.40752 kg ×385 J/kg

C×80C

⇒Q= 43086.528 J.

Therefore, approximately 43.1 kJ of heat needs to be added to increase the

temperature of the copper rod to 100

Question 23

Question

A copper block initially at a temperature of 200◦C is placed in a large ice-water

bath at 0◦C. The block’s final temperature is found to be 10◦C. The specific

heat capacity of copper is 0.385 J/g·K. If the mass of the copper block is 500 g,

calculate the amount of heat absorbed by the block during this process.

Solution

Step 1: Calculate the heat absorbed by the copper block to increase its temper-

ature from 200◦C to 10◦C using the formula:

Q=mc∆T

Where: - Qis the heat absorbed, - mis the mass of the copper block, - cis

the specific heat capacity of copper, and - ∆Tis the change in temperature.

Given that m= 500 g, c= 0.385 J/g·K, ∆T= 10 ◦C - 200 ◦C = −190 K,

we have:

Q= 500 g ×0.385 J/g ·K×(−190) K

Q=−36550 J

The negative sign indicates that heat was released by the copper block.

Step 2: Calculate the heat absorbed by the copper block when it reaches

thermal equilibrium with the ice-water bath at 0◦C. Since the final temperature

is 10◦C, the block absorbs heat as it changes from 0◦C to 10◦C. Use the same

formula as before:

Q=mc∆T

Where m= 500 g, c= 0.385 J/g·K, ∆T= 10 K.

Q= 500 g ×0.385 J/g ·K×10 K

Q= 1925 J

Step 3: The total heat absorbed by the copper block during the entire process

is the sum of the two heats calculated:

Total heat absorbed = −36550 J + 1925 J

Total heat absorbed = −34625 J

Therefore, the amount of heat absorbed by the block during the entire pro-

cess is 34625 J.

Question 24

Question

A piece of aluminum has a mass of 500 g and an initial temperature of 100

C. It

is placed in a container with 1 kg of water at a temperature of 20

C. Assuming no

heat is lost to the surroundings, what will be the final equilibrium temperature

of the system? (Specific heat capacity of aluminum = 0.9 J/g

C, specific heat

capacity of water = 4.18 J/g

Solution

Step 1: Calculate the heat lost by the aluminum and the heat gained by the

water. The heat lost by the aluminum can be calculated using the formula:

Qaluminum =m·c·∆T

where: - m= mass of the aluminum (500 g) - c= specific heat capacity of

aluminum (0.9 J/g

C) - ∆T= change in temperature of the aluminum

Since the aluminum is cooling down (100

C to the final equilibrium temper-

ature):

∆Taluminum = (final temperature) −100

The heat gained by the water can be calculated using the formula:

Qwater =m·c·∆T

where: - m= mass of the water (1000 g) - c= specific heat capacity of water

(4.18 J/g

C) - ∆T= change in temperature of the water

Since the water is heating up (final equilibrium temperature to 20

C):

∆Twater = (final temperature) −20

Step 2: Set up the heat transfer equation. Since no heat is lost to the

surroundings, the heat lost by the aluminum is equal to the heat gained by the

water:

Qaluminum =Qwater

maluminum ·caluminum ·∆Taluminum =mwater ·cwater ·∆Twater

Step 3: Solve for the final equilibrium temperature. Substitute the expres-

sions for ∆Taluminum and ∆Twater into the equation above and solve for the final

equilibrium temperature. After calculating, the final equilibrium temperature

will be 35.7C.

Question 25

Question

A metal bar of length 2 m and cross-sectional area 0.01 m2is heated from 20◦C

to 80◦C. The thermal conductivity of the metal is 400 W/(m·K), and the heat

capacity of the metal is 450 J/(kg·K). Find the rate of heat transfer through

the bar.

Solution

Step 1: Calculate the temperature difference Given that the initial temperature

(Ti) is 20◦C and the final temperature (Tf) is 80◦C, the temperature difference

(∆T) is:

∆T=Tf−Ti= 80◦C−20◦C= 60◦C= 60K

Step 2: Calculate the rate of heat transfer The rate of heat transfer can be

calculated using Fourier’s law of heat conduction, which states:

Q=−kAdT

where: Q= rate of heat transfer (W), k= thermal conductivity of the material

(W/(m·K)), A= cross-sectional area of the bar (m2), dT

dx = temperature gradient

(K/m).

Step 3: Calculate the temperature gradient The temperature gradient can

be calculated as: dT

dx =∆T

L=60K

2m= 30K/m

Step 4: Substitute values into Fourier’s law Substitute the given values into

the equation:

Q=−kAdT

dx =−(400W/(m·K))(0.01m2)(30K/m)

Step 5: Calculate the rate of heat transfer

Q=−(400)(0.01)(30) = −120W

Answer: The rate of heat transfer through the bar is 120 W.

Question 26

Question

A sample of unknown metal is heated from 25

C to 225

C. If the energy required

to raise the temperature of the metal is 450 J, and the specific heat capacity of

the metal is 0.2 J/g

C, what is the mass of the metal sample?

Solution

Step 1: Calculate the change in temperature. Given that the initial temperature

T1= 25◦C and the final temperature T2= 225◦C, we can find the change in

temperature ∆T:

∆T=T2−T1= 225◦C−25◦C = 200◦C

Step 2: Use the specific heat capacity formula to find the mass. The formula

for the energy required to change the temperature of a substance is given by:

Q=mc∆T

where: - Qis the energy required (450 J), - mis the mass of the sample (un-

known), - cis the specific heat capacity (0.2 J/g

C), and - ∆Tis the change in

temperature (200

C).

Substitute the given values into the formula and solve for m:

450 J = m×0.2 J/g

C×200

450 = 40m

m=450

40 = 11.25 g

Therefore, the mass of the metal sample is 11.25 g.

Question 27

Question

A piece of copper with a mass of 0.2 kg is heated from 20

C to 80

C. Calculate

the amount of heat energy required to raise the temperature of the copper.

(Specific heat capacity of copper is 390 J/kg◦C)

Solution

Step 1: Calculate the change in temperature of the copper. Given: Initial

temperature, T1= 20◦C Final temperature, T2= 80◦C

Change in temperature, ∆T=T2−T1∆T= 80◦C−20◦C = 60◦C

Step 2: Calculate the amount of heat energy required using the formula:

Q=mc∆T

where: Q= heat energy (in joules) m= mass of the copper (in kg) c= specific

heat capacity of copper (390 J/kg◦C) ∆T= change in temperature (in

Substitute the values into the formula:

Q= 0.2 kg ×390 J/kg◦C×60◦C

Step 3: Calculate the amount of heat energy required.

Q= 0.2×390 ×60

Q= 4680 Joules

Therefore, the amount of heat energy required to raise the temperature of

the copper from 20

C to 80

C is 4680 Joules.

Question 28

Question

A steel block of mass 2 kg is initially at a temperature of 250

C. It is then

submerged in 5 kg of water at 20

C in an insulated container. Assuming no

heat is lost to the surroundings, determine the final temperature of the system

once thermal equilibrium is reached. The specific heat capacity of steel is 450

J/kg

C and that of water is 4180 J/kg

Solution

Step 1: Calculate the heat lost by the steel block to reach thermal equilibrium.

Given that the specific heat capacity of steel is 450 J/kg

C, we can use the

formula for heat transfer:

Qsteel =mc∆T,

where mis the mass of the steel block, cis the specific heat capacity of steel,

and ∆Tis the change in temperature. Plugging in the values, we get

Qsteel = 2 ×450 ×(250 −T),

where Tis the final temperature of the system.

Step 2: Calculate the heat gained by the water to reach thermal equilibrium.

Using the same formula for heat transfer but for water this time:

Qwater =mc∆T,

where mis the mass of the water, cis the specific heat capacity of water, and

∆Tis the change in temperature. Plugging in the values, we get

Qwater = 5 ×4180 ×(T−20).

Step 3: Since there is no heat lost or gained to the surroundings, the heat

lost by the steel block is equal to the heat gained by the water:

Qsteel =Qwater.

Step 4: Equate the two expressions for heat transfer and solve for the final

temperature T:

2×450 ×(250 −T)=5×4180 ×(T−20).

Step 5: Solve for T:

900(250 −T) = 20900(T−20),

225000 −900T= 20900T−418000,

226900T= 643000,

T≈2.84

Therefore, the final temperature of the system when thermal equilibrium is

reached is approximately 2.84

Question 29

Question

A copper block of mass 0.5 kg is heated from 20

C to 100

C. The specific heat

capacity of copper is 386 J/kg

C. Calculate the heat energy required to heat

the copper block.

Solution

Step 1: Calculate the change in temperature. Given: Initial temperature, Ti=

20CFinal temperature, Tf= 100CChange in temperature, ∆T=Tf−Ti=

100C−20C= 80C

Step 2: Calculate the heat energy using the formula:

Q=mc∆T

where: Q= Heat energy m= Mass of the copper block = 0.5 kg c= Specific

heat capacity of copper = 386 J/kg

Step 3: Substitute the values into the formula and calculate.

Q= (0.5 kg)(386 J/kg

C)(80C)

Q= 19300 J

Therefore, the heat energy required to heat the copper block is 19300 J.

Question 30

Question

A block of copper with a mass of 500 g and an initial temperature of 100◦C is

dropped into 500 g of water at 20◦C. Assuming no heat is lost to the surround-

ings, what will be the final equilibrium temperature of the system? (Specific heat

capacity of copper = 0.385 J/g◦C, specific heat capacity of water = 4.184 J/g◦C)

Solution

Let the final equilibrium temperature of the system be T.

Step 1: Calculate the heat gained by copper The heat gained by

copper can be calculated using the formula:

Qcopper =mcopper ×ccopper ×(T−100◦C)

Qcopper = 500 ×0.385 ×(T−100)

Step 2: Calculate the heat lost by water The heat lost by water can

be calculated using the formula:

Qwater =mwater ×cwater ×(T−20◦C)

Qwater = 500 ×4.184 ×(T−20)

Step 3: Set up the heat exchange equation Since there is no heat lost

to the surroundings, the heat gained by the copper must be equal to the heat

lost by the water:

Qcopper =Qwater

500 ×0.385 ×(T−100) = 500 ×4.184 ×(T−20)

Step 4: Solve for the final temperature TSolving the equation from

Step 3, we get:

0.385 ×(T−100) = 4.184 ×(T−20)

0.385T−38.5=4.184T−83.68

3.799T= 45.18

T=45.18

3.799 ≈11.89◦C

Therefore, the final equilibrium temperature of the system is approximately

11.89◦C.

Question 31

Question

A copper rod of length 2 m and cross-sectional area 1 cm2is used to conduct

heat from one end to the other. The temperature at the hot end is 120◦C and at

the cold end is 40◦C. If the heat conduction along the rod is steady and uniform,

calculate the rate of heat transfer through the rod. The thermal conductivity

of copper is 390 W/(m·K).

Solution

Step 1: Calculate the temperature difference across the rod. The temperature

difference (∆T) across the rod is given by:

∆T=Thot end −Tcold end

∆T= 120◦C−40◦C = 80◦C

Step 2: Convert the temperature difference to Kelvin. To convert from

Celsius to Kelvin, we add 273 to the temperature in Celsius:

80◦C = 80 K

Step 3: Calculate the rate of heat transfer. The rate of heat transfer (Q)

through the rod can be calculated using Fourier’s law of heat conduction:

Q=−kA∆T

where: Q= rate of heat transfer, k= thermal conductivity of the material, A

= cross-sectional area of the rod, and L= length of the rod.

Substitute the given values:

Q=−390 W/(m ·K) ×1×10−4m2×80 K

2 m

Q=−390 ×1×10−4×40 = −0.156 W

Therefore, the rate of heat transfer through the rod is −0.156 W.

Question 32

Question

A copper block of mass 500 g is initially at a temperature of 100

C. It is placed

in a calorimeter containing 200 g of water at 20

C. If the final temperature of

the system is 30

C, calculate the heat capacity of the calorimeter. (Specific heat

capacity of copper = 0.39 J/g

C, specific heat capacity of water = 4.18 J/g

and heat capacity of the calorimeter is assumed to be constant and independent

of temperature.)

Solution

Step 1: Calculate the heat lost by the copper block to reach the final temperature

of 30

C. The formula to calculate heat is Q=mc∆T, where Qis the heat, mis

the mass, cis the specific heat capacity, and ∆Tis the change in temperature.

Given that the initial temperature of the copper block is 100

C and the final

temperature is 30

C, the change in temperature is ∆T= 30 −100 = −70

(negative because the temperature is decreasing).

Therefore, the heat lost by the copper block can be calculated as: Qcopper =

500 g ×0.39 J/g

C×(−70)

CQcopper =−13650 J

Step 2: Calculate the heat gained by the water and calorimeter to reach the

final temperature of 30

C. The total heat gained by the water and calorimeter

is equal to the heat lost by the copper block (by the principle of conservation

of energy).

Let Cbe the heat capacity of the calorimeter, then the total heat gained

can be expressed as: Qwater +Qcalorimeter =−13650 J

The heat gained by the water can be calculated as: Qwater = 200 g ×

4.18 J/g

C×(30 −20)

CQwater = 8360 J

Substitute this into the total heat gained equation: 8360 + Qcalorimeter =

−13650

Step 3: Calculate the heat capacity of the calorimeter. Solving for Qcalorimeter:

Qcalorimeter =−13650 −8360 Qcalorimeter =−22010 J

Since Qcalorimeter =C×∆T, and assuming the heat capacity of the calorime-

ter is constant, we can find Cas: C=Qcalorimeter

∆T=−22010 J

CC=−2201 J/

Therefore, the heat capacity of the calorimeter is 2201 J/

Question 33

Question

A piece of aluminum with mass 0.5 kg and specific heat capacity 900 J/kg◦C

is heated from an initial temperature of 20◦C to a final temperature of 60◦C.

Calculate the amount of heat energy required to raise the temperature of the

aluminum.

Solution

Step 1: Calculate the change in temperature. Given: Initial temperature (Ti)

= 20◦C, Final temperature (Tf) = 60◦C.

The change in temperature (∆T) is given by:

∆T=Tf−Ti= 60◦C−20◦C = 40◦C

Step 2: Calculate the amount of heat energy. The formula to calculate the

amount of heat energy (Q) required to raise the temperature of a substance is:

Q=mc∆T

where: m= mass of the substance, c= specific heat capacity of the substance,

∆T= change in temperature.

Given: m= 0.5 kg, c= 900 J/kg◦C, ∆T= 40◦C.

Substitute the values into the formula:

Q= (0.5 kg)(900 J/kg◦C)(40 C)

Q= 18000 J

Therefore, the amount of heat energy required to raise the temperature of

the aluminum is 18000 J.

Question 34

Question

A 2.0 kg block of copper at 100

C is dropped into 2.0 kg of water at 20

Assuming no heat is lost to the surroundings, what will be the final equilibrium

temperature of the mixture? (Specific heat of copper cCu = 390 J/kg·K, specific

heat of water cH2O = 4186 J/kg ·K, and latent heat of fusion of water Lf=

3.34 ×105J/kg)

Solution

Step 1: Determine the amount of heat lost or gained by the copper block to

reach equilibrium with the water: The specific heat equation Q=mcCu∆Tcan

be used to find the heat lost by copper. The change in temperature ∆Twill

be Tf−100, where Tfis the final temperature of the mixture. Setting up the

equation:

Qcopper =mcCu(Tf−100)

Qcopper = 2.0×390 ×(Tf−100)

Step 2: Determine the amount of heat gained by the water: The water

will absorb the heat that the copper block loses. Using the same specific heat

equation:

Qwater =mcH2O(Tf−20)

Qwater = 2.0×4186 ×(Tf−20)

Step 3: The heat lost by the copper block must equal the heat gained by the

water:

Qcopper =Qwater

2.0×390 ×(Tf−100) = 2.0×4186 ×(Tf−20)

Step 4: Solve for the final equilibrium temperature Tf:

780(Tf−100) = 8372(Tf−20)

780Tf−78000 = 8372Tf−167440

8372Tf−780Tf= 167440 −78000

7592Tf= 89440

Tf=89440

7592 ≈11.8

Therefore, the final equilibrium temperature of the mixture will be approx-

imately 11.8

Question 35

Question

A copper rod of length 2 m and cross-sectional area 0.01 m2is initially at a

temperature of 100◦C. It is then placed in an ice-water bath at 0◦C. Given that

the thermal conductivity of copper is 400 W/mK and the specific heat capacity

of copper is 390 J/kgK, calculate the time it takes for the temperature of the

rod to drop to 10◦C. Assume no heat is lost to the surroundings.

Solution

Step 1: Calculate the mass of the copper rod. Given: Length of copper rod,

L= 2 m. Cross-sectional area of copper rod, A= 0.01 m2.

The volume of the copper rod is given by:

V=A×L= 0.01 m2×2 m = 0.02 m3

The density of copper is approximately 8900 kg/m3, so the mass of the copper

rod is:

m= density ×volume = 8900 kg/m3×0.02 m3= 178 kg

Step 2: Calculate the initial heat content of the copper rod. The specific

heat capacity of copper is 390 J/kgK. The initial temperature of the copper rod

is 100◦C and the final temperature is 10◦C, so the change in temperature is:

∆T= final temperature −initial temperature = 10◦C−100◦C = −90◦C

The initial heat content of the copper rod is given by:

Q= mass×specific heat capacity×∆T= 178 kg×390 J/kgK×−90 K = −6.63×106J

Step 3: Calculate the rate of heat transfer. The thermal conductivity of

copper is 400 W/mK. The rate of heat transfer through the rod is given by

Fourier’s law: ∆Q

∆t=kA∆T

where: ∆Qis the heat transfer, ∆tis the time taken for the temperature to

drop by ∆T,kis the thermal conductivity of copper, Ais the cross-sectional

area, ∆Tis the change in temperature, and Lis the length of the rod.

Substitute the given values into the equation:

−6.63 ×106J

∆t= 400 W/mK ×0.01 m2×

−90 K

2 m

Solve for ∆tto find the time it takes for the temperature of the rod to drop

to 10◦C.

Question 2

Question

A copper rod of length 2.0 m and radius 0.5 cm is heated from an initial tem-

perature of 20

C to a final temperature of 80

C. If the coefficient of linear ex-

pansion of copper is 1.7×10−5per degree Celsius and its specific heat capacity

is 0.385 J/g

C, calculate the heat transferred to the rod. Assume the rod is

perfectly insulated and that the temperature is uniform throughout.

Solution

Step 1: Calculate the change in length of the copper rod due to heating. Given

the coefficient of linear expansion of copper, α= 1.7×10−5

C−1, the change in

length, ∆L, can be calculated using the formula:

∆L=α·L·∆T

where Lis the initial length, ∆Tis the change in temperature. Substituting

the values:

∆L= (1.7×10−5)×2.0 m ×(80 −20)

C=0.0068 m

Step 2: Calculate the volume expansion of the copper rod. The volume

expansion can be calculated using the formula for the volume of a cylinder:

V=πr2h

where ris the initial radius and his the initial length. The change in volume,

∆V, is given by:

∆V=π(r+ ∆r)2−r2L=πr2+ 2r∆r+ (∆r)2−r2L=π·2r∆r·L

Substituting the values:

∆V=π·2(0.005 m)(0.0068 m) ·2 m = 0.00043 m3

Step 3: Calculate the heat transferred to the rod. The heat transferred can

be calculated using the formula:

Q=mc∆T

where mis the mass of the copper rod, cis the specific heat capacity of cop-

per, and ∆Tis the change in temperature. Since the density of copper is

8,960 kg/m3, the mass of the rod can be calculated as:

m=ρV = 8960 kg/m3×0.00043 m3= 3.848 kg

Substitute the values to find the heat transferred:

Q= 3.848 kg ×0.385 J/g

C×(80 −20)

C = 355.08 J

Question 3

Question

An aluminum rod of length 1 m and cross-sectional area 0.01 m2is heated from

20◦C to 80◦C. The rod has a thermal conductivity of 200 W/(m

K), a specific

heat capacity of 900 J/(kg

K), and a density of 2700 kg/m3. Calculate the

amount of heat transferred to the rod during this process.

Solution

Step 1: Calculate the mass of the aluminum rod. Given the density of aluminum

is 2700 kg/m3and the volume of the rod is V= 1 m ×0.01 m2= 0.01 m3, the

mass mcan be calculated as:

m= density ×volume = 2700 kg/m3×0.01 m3= 27 kg

Step 2: Calculate the change in temperature. The change in temperature

∆T=Tf−Ti= 80◦C−20◦C = 60 K.

Step 3: Calculate the amount of heat transferred. The amount of heat

transferred Qis given by the formula:

Q=mc∆T

Substitute the mass m= 27 kg, specific heat capacity c= 900 J/(kg

K), and

change in temperature ∆T= 60 K:

Q= 27 kg ×900 J/(kg

K) ×60 K = 1458000 J

Therefore, the amount of heat transferred to the aluminum rod during this

process is 1,458,000 J.

Question 4

Question

A copper rod of length 1.5 m and cross-sectional area 4 cm2is heated from

C to 80

C. If the coefficient of linear expansion of copper is 1.7×10−5

C−1

and the specific heat capacity of copper is 0.387 J/g

C, calculate the change in

length of the rod and the heat absorbed by the rod.

Solution

Step 1: Calculate the change in length of the rod due to heating. Given the

initial temperature T1= 20

C, final temperature T2= 80

C, and coefficient

of linear expansion α= 1.7×10−5

C−1, the change in length is given by the

formula:

∆L=L·α·∆T

where ∆T=T2−T1and Lis the original length of the rod.

Substitute the given values:

∆L= 1.5 m ·1.7×10−5

C−1·(80 −20)

∆L= 1.5×10−2m=0.015 m

Step 2: Calculate the heat absorbed by the rod. The heat absorbed by the

rod can be calculated using the formula:

Q=mc∆T

where mis the mass of the rod, cis the specific heat capacity of copper, and

∆Tis the change in temperature.

First, calculate the mass of the rod using the formula for volume:

V=A·L

m=ρ·V

where ρ= 8.96 g/cm3is the density of copper.

Substitute the values to find the mass m:

m= 8.96 g/cm3·4 cm2·1.5 m ·104cm2/m2

m= 538.56 kg = 538560 g

Now, substitute the mass m= 538560 g, specific heat capacity c= 0.387

J/g

C, and change in temperature ∆T= 80 −20 = 60

C into the formula:

Q= 538560 g ·0.387 J/g◦C·60

Q= 12324480 J

Therefore, the change in length of the rod is 0.015 m and the heat absorbed

by the rod is 12324480 J.

Question 5

Question

A copper rod of length 2 meters and diameter 1 cm is heated from 20

C to

C. If the thermal conductivity of copper is 401 W/(m.K), calculate the heat

conducted through the rod during this process.

Solution

Let’s first calculate the surface area of the rod. The formula for the surface area

of a cylinder is given by:

A= 2πrh + 2πr2

where ris the radius of the rod and his the length of the rod.

Step 1: Calculate the radius of the rod. Given that the diameter of the rod

is 1 cm, the radius ris half of the diameter.

r=1

2×(1 cm) = 0.5 cm = 0.005 m

Step 2: Calculate the surface area of the rod.

A= 2π×0.005 ×2+2π×0.0052

A= 0.063 m2

Step 3: Calculate the temperature difference. The temperature difference

is 80 ◦C−20 ◦C = 60 ◦C = 60 K.

Step 4: Calculate the heat transfer. The formula for heat transfer is given

by:

Q=k·A·∆T

where Qis the heat transfer, kis the thermal conductivity, Ais the surface

area, and ∆Tis the temperature difference.

Substitute the values into the formula:

Q= 401 W/(m.K) ×0.063 m2×60 K

Q= 1519.4 W

Therefore, the heat conducted through the rod during this process is 1519.4

Watts.

Question 6

Question

A copper cylinder with a radius of 5 cm and a height of 10 cm is initially at a

temperature of 100

C. It is then placed in a water bath at 20

C. If the specific

heat capacity of copper is 0.386 J/g

C, how much heat is transferred to the water

bath when the copper cylinder reaches thermal equilibrium with the water?