PHYS 232 - UNIVERSITY PHYSICS II Temperature and heat Question Bank Set 3

PHYS 232 - UNIVERSITY PHYSICS

II - Temperature and heat

Question Bank - Set 3

Liberty University

Question 1

Question

A copper sphere of radius 10 cm at a temperature of 400 K is placed in a large

body of water at 300 K. If the thermal conductivity of copper is 390 W/(m K),

how long does it take for the sphere to cool to 350 K?

Solution

Step 1: The rate of heat transfer from the sphere to the water can be calculated

using Newton’s Law of Cooling:

dt =kA∆T

where: dQ/dt is the rate of heat transfer, kis the thermal conductivity of copper,

Ais the surface area of the sphere, and ∆Tis the temperature difference between

the sphere and the water.

Step 2: The surface area of a sphere is 4πr2, where ris the radius of the

sphere. Given r= 10 cm, the surface area A= 4π(0.1)2.

Step 3: The temperature difference is ∆T=T1−T2where T1= 400 K and

T2= 300 K. Thus, ∆T= 400 −300.

Step 4: Substituting the values into the formula from Step 1:

dt = 390 ×4π(0.1)2×(400 −300)

Step 5: Simplifying the expression gives:

dt = 390 ×4π(0.01) ×100

Step 6: Solving for dQ/dt gives:

dt = 124.32πW

Step 7: To further solve for the time it takes for the sphere to cool to 350

K, we need to evaluate the heat lost until the sphere cools to that temperature.

Let Qbe the heat lost, and mbe the mass of the sphere, ρbe the density of

copper, and cbe the specific heat capacity of copper.

Step 8: The heat lost is given by:

Q=mc∆T

Step 9: Since Q=RdQ, we have:

mc∆T=ZdQ

dt dt

Step 10: Integrating both sides gives:

mc∆T=Z124.32π dt

Step 11: The limits of integration are from 0 to t. This results in:

mc∆T= 124.32πt

Step 12: Substituting known values for m,ρ,c,r, and solving for tgives the

time required for the sphere to cool to 350 K.

Question 2

Question

A 500 g block of copper at 300 K is dropped into 2.5 L of water at room

temperature (20

C). Assuming no heat is lost to the surroundings, what is the

final temperature of the system? (Specific heat capacity of copper = 385 J/kg·K,

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

Solution

Step 1: Calculate the initial heat energy of the copper block: The formula for

heat energy (Q) is given by Q=mc∆T, where: - mis the mass of the object, -

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

Substitute m= 0.5 kg, c= 385 J/kg ·K, and ∆T=Tfinal −300 K into the

formula to get: Qcopper = (0.5 kg)(385 J/kg ·K)(Tfinal −300 K)

Step 2: Calculate the initial heat energy of the water: Using the same

formula, we get: Qwater = (m)(c)(∆T) Substitute m= 2.5 kg, c= 4186 J/kg·K,

and ∆T=Tfinal −293 K to get: Qwater = (2.5 kg)(4186 J/kg ·K)(Tfinal −293 K)

Step 3: Since no heat is lost to the surroundings, Qcopper =−Qwater

Step 4: Equate the two heat energies and solve for Tfinal: (0.5 kg)(385 J/kg ·

K)(Tfinal −300 K) = −(2.5 kg)(4186 J/kg ·K)(Tfinal −293 K)

Solve for Tfinal to find the final temperature of the system.

Question 3

Question

A copper bar of length 1 m and cross-sectional area 1 cm2is heated to 100◦C

and then placed in a room at 20◦C. The thermal conductivity of copper is

390 W/mK and its specific heat capacity is 390 J/kgK. Assuming no heat loss

to the surroundings, calculate the time it takes for the bar to cool to 30◦C.

Solution

Step 1: Calculate the mass of the copper bar.

Given that the length of the bar is 1 m, the cross-sectional area is 1 cm2=

1×10−4m2. The volume of the bar is then 1 m ×1×10−4m2= 10−4m3.

Since the density of copper is 8900 kg/m3, the mass of the bar is 10−4m3×

8900 kg/m3= 0.89 kg.

Step 2: Calculate the heat capacity of the copper bar.

The heat capacity Cof a material is given by C= mass×specific heat capacity.

Substituting in the values, we find C= 0.89 kg ×390 J/kgK = 347.1 J/K.

Step 3: Calculate the initial heat energy in the bar.

The initial temperature of the bar is 100◦C = 373 K. So, the initial heat energy

Q1in the bar is given by Q1=C×∆T= 347.1 J/K ×(373 −293) K = 27768 J.

Step 4: Calculate the final heat energy in the bar.

The final temperature of the bar is 30◦C = 303 K. So, the final heat energy Q2

in the bar is given by Q2=C×∆T= 347.1 J/K ×(303 −293) K = 3471 J.

Step 5: Calculate the rate of heat transfer through the bar.

The rate of heat transfer Pthrough a material of thermal conductivity κis

given by P=κ×A×∆T

L. Substituting the values, we find P= 390 W/mK ×

1×10−4m2×373−303

1K = 0.78 W.

Step 6: Calculate the time taken for the bar to cool.

The rate of heat transfer is also given by P=Q1−Q2

t, where tis the time taken.

Substituting the values, we have 0.78 = 27768−3471

t, which gives t=24363

0.78 =

31231 s = 8.67 hours.

Therefore, it takes approximately 8.67 hours for the copper bar to cool from

100◦C to 30◦C.

Question 4

Question

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

C is

dropped into a container of water at 20

C. If the final temperature of the system

is 25

C, determine the mass of the water in the container. Assume no heat is

lost to the surroundings and the specific heat capacities of copper and water are

0.39 J/g

C and 4.18 J/g

C, respectively.

Solution

Let mwbe the mass of water in the container.

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

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

Qwater =mw×cwater ×∆T

where mw= mass of water, cwater = specific heat capacity of water, ∆T=

Tfinal −Tinitial.

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

Qcopper =mcopper ×ccopper ×∆T

where mcopper = mass of copper, ccopper = specific heat capacity of copper.

Given that the final temperature is 25

C and the initial temperature of the

copper block is 100

C, and the initial temperature of the water is 20

C, we have:

∆Twater = 25C−20C= 5C

∆Tcopper = 25C−100C=−75C

Substitute the values into the formulas to find the heat gained by the water

and the copper block.

Step 2: Set up the equation based on the conservation of energy. According

to the conservation of energy, the heat lost by the copper block should be equal

to the heat gained by the water.

Qcopper =Qwater

Step 3: Solve for the mass of water. Substitute the equations for Qcopper

and Qwater from Step 1 into the equation in Step 2. Solve for mwto find the

mass of water in the container.

Question 5

Question

A copper block of mass 0.5 kg at an initial temperature of 100

C is placed in

a container with 1 kg of water at 20

C. If the final temperature of the system

is 30

C, calculate the heat gained or lost by the copper block. Assume specific

heat capacities: cCu = 385 J/kg

C and cH2O = 4186 J/kg

Solution

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

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

QCu =mCu ·cCu ·∆TCu

Where: - QCu is the heat lost by the copper block, - mCu = 0.5 kg is the

mass of the copper block, - cCu = 385 J/kg

C is the specific heat capacity of

copper, and - ∆TCu =Tf−Ti= 30C−100C=−70C(Negative because the

temperature is decreasing).

Substitute the values into the formula:

QCu = 0.5 kg ×385 J/kg

C×(−70C)

QCu =−13,475 J

Therefore, the copper block loses 13,475 J of heat.

Step 2: Calculate the heat gained by the water. The heat gained by the

water can be calculated using the formula:

QH2O =mH2O ·cH2O ·∆TH2O

Where: - QH2O is the heat gained by the water, - mH2O = 1 kg is the mass

of the water, - cH2O = 4186 J/kg

C is the specific heat capacity of water, and -

∆TH2O =Tf−Ti= 30C−20C= 10C

Substitute the values into the formula:

QH2O = 1 kg ×4186 J/kg

C×10C

QH2O = 41,860 J

Therefore, the water gains 41,860 J of heat.

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

the heat lost by the copper block must be equal to the heat gained by the water.

Thus, the heat lost by the copper block is equal in magnitude but opposite in

sign to the heat gained by the water.

Hence, the heat gained or lost by the copper block is 13,475 J of heat.

Question 6

Question

A 500 g block of copper at an initial temperature of 100

C is dropped into 1 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

= 0.385 J/g

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

C).

Solution

Step 1: Calculate the heat lost by the copper block. We can use the formula

Q=mc∆T

where: - Qis the heat lost or gained, - mis the mass of the substance, - cis the

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

Given: - m= 500 g, - c= 0.385 J/g

C, - ∆T=Tfinal −Tinitial =Tfinal −100

Therefore, the heat lost by the copper block is:

Qcopper = 500 ×0.385 ×(Tfinal −100)

Step 2: Calculate the heat gained by the water. Using the same formula as

above, with: - m= 1000 g, - c= 4.18 J/g

C, - ∆T=Tfinal −20

The heat gained by the water is:

Qwater = 1000 ×4.18 ×(Tfinal −20)

Since no heat is lost to the surroundings, we have:

Qcopper =Qwater

500 ×0.385 ×(Tfinal −100) = 1000 ×4.18 ×(Tfinal −20)

Step 3: Solve for the final equilibrium temperature (Tfinal). After simplifying

and solving the equation, we find:

Tfinal = 32.8

Therefore, the final equilibrium temperature of the system is 32.8

Question 7

Question

A solid copper sphere of radius 5 cm is heated until its temperature is increased

by 100

C. If the specific heat capacity of copper is 0.386 J/g

C and its density is

8.96 g/cm3, calculate the heat energy supplied to the sphere during this process.

Solution

Step 1: Calculate the mass of the copper sphere. Given that the density of

copper is 8.96 g/cm3, the volume of the sphere can be calculated using the

formula for the volume of a sphere:

V=4

3πr3.

Substituting r= 5 cm into the formula:

V=4

3π(5 cm)3.

V=500

3πcm3.

Since the density is 8.96 g/cm3, the mass mof the sphere can be calculated

using the formula: m= density ×volume.Substituting the values:

m= 8.96 ×500

3πg.

Step 2: Calculate the heat energy supplied. The heat energy supplied can

be calculated using the formula: Q=mc∆T, where: - Qis the heat energy

supplied, - mis the mass of the sphere, - cis the specific heat capacity of

copper, and - ∆Tis the change in temperature. Substitute known values:

Q= 8.96 ×500

3π×0.386 ×100 J.

Q= 46.939πJ.

Question 8

Question

A copper ball of mass 0.2 kg at a temperature of 100

C is dropped into a

calorimeter containing 0.5 kg of water at 20

C. If the final equilibrium tem-

perature of the system is 25

C, calculate the specific heat capacity of copper.

Assume no heat is lost to the surroundings.

Solution

Step 1: Calculate the heat lost by the copper ball. The heat lost by the copper

ball can be calculated using the formula:

Qcopper =mc∆T

where: - m= 0.2 kg is the mass of the copper ball, - cis the specific heat

capacity of copper (to be determined), - ∆T= 100 −25 = 75

C is the change

in temperature.

Plugging in the values:

Qcopper = 0.2×c×75

Step 2: Calculate the heat gained by the water in the calorimeter. The heat

gained by the water in the calorimeter can be calculated using the formula:

Qwater =mc∆T

where: - m= 0.5 kg is the mass of the water, - c= 4186 J/kg

C is the specific

heat capacity of water, - ∆T= 25 −20 = 5

C is the change in temperature.

Plugging in the values:

Qwater = 0.5×4186 ×5

Step 3: Set up the heat lost equals heat gained equation. Since there is no

heat transfer to the surroundings, the heat lost by the copper ball is equal to

the heat gained by the water:

Qcopper =Qwater

Step 4: Solve for the specific heat capacity of copper. Setting the two

equations equal:

0.2×c×75 = 0.5×4186 ×5

Solve for c:

c=0.5×4186 ×5

0.2×75

c=10465

c= 697.67 J/kg

Therefore, the specific heat capacity of copper is 697.67 J/kg

Question 9

Question

A copper cylinder with a mass of 0.5 kg and a specific heat capacity of 385

J/(kg·K) is heated from an initial temperature of 25

C to a final temperature

of 70

C. How much heat is needed to achieve this temperature change?

Solution

Step 1: Calculate the temperature change using the formula ∆T=Tf−Ti.

∆T= 70◦C−25◦C = 45◦C

Step 2: Calculate the heat required using the formula Q=mc∆T, where: -

mis the mass of the copper cylinder (0.5 kg), - cis the specific heat capacity of

copper (385 J/(kg·K)), - ∆Tis the temperature change (45

C).

Q= 0.5 kg ×385 J/(kg ·K) ×45 K

Q= 8662.5 J

Therefore, the amount of heat needed to achieve the temperature change is

8662.5 J.

Question 10

Question

A 0.5 kg block of ice at -10

C is placed in a room at 20

C. The block of ice

absorbs heat until it reaches its melting point. Calculate the amount of heat

absorbed by the ice during this process. The specific heat capacity of ice is

2090 J/kg ·

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

C, and the

latent heat of fusion of ice is 334 J/g.

Solution

Step 1: The heat absorbed by the ice to reach its melting point can be calculated

as follows: When the ice is heated from -10

C to 0

C: Mass of the ice, m= 0.5 kg

Specific heat capacity of ice, cice = 2090 J/kg·

C Change in temperature, ∆T=

0C−(−10C) = 10CThe heat absorbed is given by the formula Q=mcice∆T.

Plugging in the values, we get:

Q= 0.5 kg ·2090 J/kg ·

C·10C= 10450 J

Step 2: The ice melts at 0

C, and the heat absorbed is used to change the

ice into water at 0

C. The heat required for this phase change is given by: Mass

of the ice, m= 0.5 kg = 500 g Latent heat of fusion of ice, L= 334 J/g The heat

absorbed during this phase change is Q=mL. Plugging in the values, we get:

Q= 500 g ·334 J/g = 167000 J

Step 3: The water at 0

C is heated until it reaches 20

C. The heat absorbed

during this temperature change can be calculated as follows: Mass of the water,

m= 0.5 kg Specific heat capacity of water, cwater = 4186 J/kg ·

C Change in

temperature, ∆T= 20C−0C= 20CThe heat absorbed is given by the formula

Q=mcwater ∆T. Plugging in the values, we get:

Q= 0.5 kg ·4186 J/kg ·

C·20C= 41860 J

Step 4: The total heat absorbed by the ice during this process is the sum

of the heats calculated in steps 1, 2, and 3: Total heat absorbed = 10450 J +

167000 J + 41860 J = 219310 J

Therefore, the amount of heat absorbed by the ice during this process is

219310 J.

Question 11

Question

A copper vessel contains 0.5 kg of water at a temperature of 20

C. If 0.1 kg

of ice at -10

C is added to the water, calculate the final temperature of the

system after thermal equilibrium is achieved. (Specific heat capacity of water

= 4200 J/kg◦C, specific heat capacity of copper = 386 J/kg◦C, latent heat of

fusion of ice = 334 J/g)

Solution

Step 1: Calculate the heat gained or lost by the water to reach the final tem-

perature.

The heat gained by the water is given by:

Qwater =mwater ·cwater ·(Tf−Twater, initial)

where: - mwater = 0.5 kg (mass of water), - cwater = 4200 J/kg◦C (specific

heat capacity of water), - Twater, initial = 20◦C, - Tf= final temperature of the

system.

Step 2: Calculate the heat gained or lost by the ice to reach the final tem-

perature.

Considering the ice will melt initially, the heat gained by the ice for the

temperature to rise from -10

C to 0

C and then to melt completely is given by:

Qice =Qtemperature change +Qfusion

where:

Qtemperature change =mice ·cice ·(0 −(−10))

Qfusion =mice ·Lf

where: - mice = 0.1 kg (mass of ice), - cice = 2100 J/kg◦C (specific heat

capacity of ice), - Lf= 334 J/g = 334000 J/kg (latent heat of fusion of ice).

Step 3: Set the total heat gained by the system to zero for thermal equilib-

rium to find the final temperature.

Qwater +Qice = 0

Substitute the results from Step 1 and Step 2 to find the final temperature

Tf.

Question 12

Question

A copper sphere of radius 5 cm is initially at a temperature of 200 K. It is

placed in a large bath of liquid nitrogen at a temperature of 77 K, and the

entire system is well-insulated. Assuming the specific heat capacity of copper is

0.385 J/g·K, the specific heat capacity of liquid nitrogen is 0.98 J/g·K, and the

density of copper is 8.96 g/cm3, determine the final temperature of the system

after thermal equilibrium is reached.

Solution

Step 1: First, we find the mass of the copper sphere. The volume of a sphere is

given by the formula V=4

3πr3, where ris the radius.

V=4

3π(5 cm)3=500

3πcm3

Given the density of copper is 8.96 g/cm3, we can calculate the mass:

mass = density ×volume = 8.96 ×500

3π≈4717.8 g

Step 2: Next, we determine the heat lost by the copper sphere and the

heat gained by the liquid nitrogen in order to set up the equation for thermal

equilibrium.

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

Qcopper =mcopper ×ccopper ×∆Tcopper

where: - mcopper is the mass of copper, - ccopper is the specific heat capacity

of copper, and - ∆Tcopper is the change in temperature of the copper sphere.

Given that the initial temperature of the copper sphere is 200 K and the

final temperature of the system is TK, we have:

∆Tcopper =T−200

Substitute the values into the formula:

Qcopper = 4717.8×0.385 ×(T−200)

The heat gained by the liquid nitrogen can be calculated similarly:

Qnitrogen =mnitrogen ×cnitrogen ×∆Tnitrogen

where mnitrogen is the mass of nitrogen and cnitrogen is the specific heat

capacity of nitrogen. Since the entire bath of nitrogen is at 77 K and the final

temperature of the system is TK, we have:

∆Tnitrogen =T−77

Substitute the values into the formula:

Qnitrogen =mnitrogen ×0.98 ×(T−77)

Step 3: At thermal equilibrium, the heat lost by the copper sphere is equal

to the heat gained by the liquid nitrogen.

Qcopper =Qnitrogen

Substitute the expressions for Qcopper and Qnitrogen:

4717.8×0.385 ×(T−200) = mnitrogen ×0.98 ×(T−77)

Step 4: Substitute mnitrogen in terms of the volume of the nitrogen bath:

mnitrogen = densitynitrogen ×volumenitrogen

The volume of the nitrogen bath (Vnitrogen) can be approximated by consid-

ering it as a large cylindrical container:

Vnitrogen =πr2h

Given that the radius of the sphere is 5 cm, the height of the cylindrical

container can be estimated as 3 times the radius of the sphere to ensure the

entire sphere is submerged:

h≈3×2×5 cm = 30 cm

Question 13

Question

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

temperature of 300 K. It is then heated until its temperature reaches 450 K. If

the rod has a specific heat capacity of 385 J/kg·K and a density of 8960 kg/m3,

calculate the amount of heat energy transferred to the rod during this process.

Solution

Step 1: Calculate the mass of the copper rod. Given: Length of the rod, L= 2

m Cross-sectional area of the rod, A= 0.01 m2Density of copper, ρ= 8960

kg/m3

The volume of the rod is given by V=A×L= 0.01 m2×2 m = 0.02 m3.

The mass of the rod can be calculated as m=ρ×V= 8960 kg/m3×

0.02 m3= 179.2 kg.

Step 2: Calculate the change in temperature. Initial temperature, Ti= 300

K Final temperature, Tf= 450 K

The change in temperature, ∆T=Tf−Ti= 450 K −300 K = 150 K.

Step 3: Calculate the amount of heat energy transferred. Specific heat

capacity of copper, c= 385 J/kg·K

The amount of heat energy transferred, Q, can be calculated using the for-

mula:

Q=mc∆T

Substitute the known values:

Q= 179.2 kg ×385 J/kg ·K×150 K

Q= 10435200 J

Therefore, the amount of heat energy transferred to the copper rod during

this process is 10,435,200 J.

Question 14

Question

An insulated container initially holds 2 kg of water at 30

C. A 1 kg block of

copper at 100

C is placed in the water and the system is allowed to reach thermal

equilibrium. Assuming no heat is lost to the surroundings, calculate the final

temperature of the system. (Specific heat capacity of water = 4186 J/kg

specific heat capacity of copper = 386 J/kg

Solution

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

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

m1c1(Tf−Tw,i) = m2c2(Tc,i −Tf)

where m1is the mass of water, c1is the specific heat capacity of water, Tf

is the final temperature, Tw,i is the initial temperature of water, m2is the

mass of copper, c2is the specific heat capacity of copper, and Tc,i is the initial

temperature of copper. Substitute the known values into the equation:

2×4186 ×(Tf−30) = 1 ×386 ×(100 −Tf)

Step 2: Solve the equation to find the final temperature.

8372(Tf−30) = 386(100 −Tf)

8372Tf−251160 = 38600 −386Tf

8372Tf+ 386Tf= 38600 + 251160

8758Tf= 289760

Tf=289760

8758 ≈33.05

Therefore, the final temperature of the system is approximately 33.05

Question 15

Question

A copper bowl of mass 0.5 kg contains 0.2 kg of water at an initial temperature

of 10

C. A piece of ice at 0

C is added to the bowl, causing the temperature

of the water to decrease to 0

C. Assuming no heat is lost to the surroundings,

calculate the mass of ice added to the bowl.

Given specific heat capacities: - Copper: ccopper = 0.385 kJ/kg◦C - Water:

cwater = 4.18 kJ/kg◦C - Latent heat of fusion of ice: 334 kJ/kg

Solution

Step 1: Calculate the heat lost by the water as it cools from 10

C to 0

C. The

heat lost is given by the formula:

Qwater =mcwater∆T

where - m= 0.2 kg is the mass of water, - cwater = 4.18 kJ/kg◦C is the specific

heat capacity of water, and - ∆T= 10C−0C= 10Cis the temperature change.

Plugging in the values gives:

Qwater = 0.2×4.18 ×10 = 8.36 kJ

Step 2: Calculate the heat absorbed by the ice to melt at 0

C. The heat

absorbed is given by the formula:

Qice =mlf

where - lf= 334 kJ/kg is the latent heat of fusion of ice, and - mis the mass of

ice that melted.

Since the water cools to 0

C, it absorbs the heat lost by the water and the

heat needed to melt the ice:

Qwater =Qice

8.36 = m×334

m=8.36

334 = 0.025 kg

Therefore, the mass of ice added to the bowl is 0.025 kg.

Question 16

Question

A steel plate with a mass of 2.5 kg and a specific heat capacity of 450 J/kg·

is placed in an oven at a temperature of 200

C. After 10 minutes, the plate

reaches thermal equilibrium with the oven at 250

C. Calculate the amount of

heat transferred to the plate during this time.

Solution

Step 1: Calculate the initial temperature of the steel plate. The initial temper-

ature of the steel plate is 200

Step 2: Calculate the final temperature of the steel plate. The final temper-

ature of the steel plate is 250

Step 3: Calculate the change in temperature of the steel plate. ∆T=

250C−200C= 50C

Step 4: Calculate the amount of heat transferred to the steel plate. The

amount of heat transferred can be calculated using the formula:

Q=mc∆T

where: - Qis the heat transferred, - mis the mass of the steel plate (2.5 kg), -

cis the specific heat capacity of steel (450 J/kg·

C), and - ∆Tis the change in

temperature (50

C).

Plugging in the values, we get:

Q= 2.5 kg ×450 J/kg ·

C×50

Q= 56250 J

Therefore, the amount of heat transferred to the plate during this time is

56250 J.

Question 17

Question

A copper kettle of mass 1.5 kg contains 2 kg of water. The temperature of the

kettle is 60

C and the water is 20

C. The copper kettle is placed on a stove and

heat is supplied until the temperature of the water reaches 100

C. Assuming no

heat is lost to the surroundings, calculate the amount of heat supplied to the

copper kettle.

Solution

Step 1: Find the heat absorbed by the water to reach 100

C. The specific heat

capacity of water, cwater, is 4186 J/kg

C. The specific heat capacity of copper,

ccopper, is 390 J/kg

The initial temperature of the water, Tinitial,water = 20

C The final tem-

perature of the water, Tfinal,water = 100

C The mass of the water, mwater = 2

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

Qwater =mwater ·cwater ·(Tfinal,water −Tinitial,water)

Qwater = 2 ·4186 ·(100 −20)

Qwater = 2 ·4186 ·80

Qwater = 669760 J

Step 2: Find the heat absorbed by the copper kettle to reach 100

C. The

initial temperature of the copper kettle, Tinitial,copper = 60

C The final temper-

ature of the copper kettle, Tfinal,copper = 100

C The mass of the copper kettle,

mcopper = 1.5 kg

The heat absorbed by the copper kettle is given by the formula:

Qcopper =mcopper ·ccopper ·(Tfinal,copper −Tinitial,copper)

Qcopper = 1.5·390 ·(100 −60)

Qcopper = 1.5·390 ·40

Qcopper = 23400 J

Step 3: Calculate the total heat supplied to the copper kettle. The total heat

supplied is the sum of the heat absorbed by the water and the heat absorbed

by the copper kettle:

Qtotal =Qwater +Qcopper

Qtotal = 669760 + 23400

Qtotal = 693160 J

Therefore, the amount of heat supplied to the copper kettle is 693160 J.

Question 18

Question

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

temperature of 150◦C. If it absorbs 5000 J of heat, what is the final temperature

of the rod? Assume the specific heat capacity of copper is 386 J/(kg·K) and its

density is 8960 kg/m3.

Solution

Step 1: Calculate the mass of the copper rod. Given that the density of copper

is 8960 kg/m3and the cross-sectional area of the rod is 4 cm2(or 0.0004 m2),

we can find the mass as follows:

Volume = Cross-sectional area ×Length = 0.0004 m2×2 m = 0.0008 m3

Mass = Volume ×Density = 0.0008 m3×8960 kg/m3= 7.168 kg

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

absorbed by the rod is 5000 J, and the specific heat capacity of copper is 386

J/(kg·K). We can calculate the change in temperature using the formula:

Q=mc∆T

where Qis the heat absorbed, mis the mass, cis the specific heat capacity, and

∆Tis the change in temperature. Rearranging the formula gives:

∆T=Q

mc =5000 J

7.168 kg ×386 J/(kg ·K) ≈1.921 K

Step 3: Calculate the final temperature of the copper rod. The initial tem-

perature of the rod is 150◦C. Therefore, the final temperature can be found by

adding the change in temperature calculated in the previous step:

Final temperature = 150 + 1.921 ≈151.92◦C

Therefore, the final temperature of the copper rod is approximately 151.92◦C.

Question 19

Question

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

C to 100

If the coefficient of linear expansion of copper is 1.7×10−5per degree Celsius,

find the increase in length of the rod.

Solution

Step 1: Calculate the initial length of the rod using the formula for the circum-

ference of a circle:

Initial length = π×diameter

Initial length = π×0.02 m = 0.0628 m

Step 2: Calculate the increase in temperature:

∆T= 100C −20C = 80C

Step 3: Calculate the change in length using the formula for linear expansion:

∆L=α×L×∆T

where αis the coefficient of linear expansion, Lis the initial length, and ∆Tis

the change in temperature.

∆L= 1.7×10−5×0.0628 ×80

∆L= 0.000085 m

Therefore, the increase in length of the rod is 0.000085 m.

Question 20

Question

A copper pot of mass 1 kg contains 0.5 kg of water at 20

C. If a piece of ice at

−10Cwith a mass of 0.2 kg is added to the pot, calculate the final temperature

of the system. Assume specific heat capacities: copper = 0.385 J/g

C, water =

4.18 J/g

C, ice = 2.09 J/g

C and the latent heat of fusion of ice = 3.33 J/g.

Solution

Step 1: Calculate the heat gained or lost by each component of the system:

Heat lost by copper = mcopper ·ccopper ·(Tf−Tinitial)

= (1 kg) ·(0.385 J/g

C) ·(Tf−20C)

= 385 J/

C·(Tf−20C)

Heat gained by ice for phase change = mice ·Lfusion

= (0.2 kg) ·(3.33 J/g)

= 0.666 kJ

Heat gained by ice to reach final temperature = mice ·cice ·(Tf−Tinitial, ice)

= (0.2 kg) ·(2.09 J/g

C) ·(Tf−(−10C))

= 0.418 kJ/

C·(Tf+ 10C)

Heat gained by water = mwater ·cwater ·(Tf−Tinitial)

= (0.5 kg) ·(4.18 J/g

C) ·(Tf−20C)

= 2.09 kJ/

C·(Tf−20C)

Question 21

Question

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

C is placed in a calorimeter

containing 1 kg of water at 20

C. If the final temperature of the system is 22

determine the specific heat capacity of copper. Assume no heat is lost to the

surroundings.

Solution

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

heat gained by the water. The heat lost by the copper block can be calculated

using the formula:

Qcopper =mc∆T

where: - m= 500 g = 0.5 kg (mass of copper block), - cis the specific heat

capacity of copper (in J/kg

C), - Tinitial = 100C, - Tfinal = 22C, - ∆T=

Tinitial −Tfinal = 100C−22C= 78C.

Substitute the values into the formula to find Qcopper.

Step 2: Next, we calculate the heat gained by the water in the calorimeter

using the formula:

Qwater =mc∆T

where: - m= 1 kg (mass of water), - c= 4186 J/kg

C (specific heat capacity of

water), - Tinitial = 20C, - Tfinal = 22C, - ∆T=Tfinal −Tinitial = 22C−20C= 2C.

Substitute the values into the formula to find Qwater.

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

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

Therefore:

Qcopper =Qwater

Step 4: Set the two expressions for heat equal to each other and solve for

the specific heat capacity of copper (c).

mc∆T=mc∆T

Substitute the calculated values of Qcopper and Qwater into the equation and

solve for c.

Question 22

Question

A steel beam of mass 100 kg at a temperature of 100

C is placed in a pool

of water at 20

C. If the specific heat capacity of steel is 450 J/kg

C and that

of water is 4186 J/kg

C, calculate the final temperature of the system when

thermal equilibrium is reached. Assume no heat is lost to the surroundings.

Solution

Step 1: Calculate the heat gained by the steel beam.

Heat gained by steel = mass ×specific heat capacity ×∆temperature

Heat gained by steel = 100 kg×450 J/kg

C×(final temperature−100C) (since steel is cooling)

Step 2: Calculate the heat lost by the water.

Heat lost by water = mass ×specific heat capacity ×∆temperature

Heat lost by water = 100 kg×4186 J/kg

C×(20C−final temperature) (since water is heating)

Step 3: Set up the heat gained by steel equal to the heat lost by water to

find the final temperature.

100 ×450 ×(final temperature −100) = 100 ×4186 ×(20 −final temperature)

Step 4: Solve for the final temperature.

45000(final temperature −100) = 418600(20 −final temperature)

45000final temperature −4500000 = 8372000 −41860final temperature

86860final temperature = 12872000

final temperature ≈148.4C

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

reached is approximately 148.4

Question 23

Question

A copper block of mass 0.5 kg is heated to a temperature of 200◦C and then

dropped into 2 kg of water at 20◦C in an insulated container. If the final tem-

perature of the system is Tafter thermal equilibrium is achieved, calculate the

value of T. Assume specific heat capacities of copper and water are 400 J/kg◦C

and 4200 J/kg◦C, respectively, and neglect any heat losses to the environment.

Solution

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

The heat lost by the copper block is given by:

Qlost =mc∆T

where: - mis the mass of the copper block (0.5 kg), - cis the specific heat

capacity of copper (400 J/kg◦C), - ∆Tis the change in temperature (from

200◦C to T).

Therefore,

Qlost = 0.5×400 ×(T−200)

The heat gained by the water is given by:

Qgained =mc∆T

where: - mis the mass of the water (2 kg), - cis the specific heat capacity of

water (4200 J/kg◦C), - ∆Tis the change in temperature (from Tto 20◦C).

Therefore,

Qgained = 2 ×4200 ×(20 −T)

At thermal equilibrium, the heat lost must equal the heat gained. Thus, we

can set up the equation:

0.5×400 ×(T−200) = 2 ×4200 ×(20 −T)

Step 2: Solve for the final temperature T.

Solving the equation from step 1, we get:

200(T−200) = 8400(20 −T)

200T−40000 = 168000 −8400T

200T+ 8400T= 168000 + 40000

8600T= 208000

T=208000

8600

Therefore, the final temperature Tafter thermal equilibrium is 208000

8600 ≈

24.19◦C.

Question 24

Question

A metal rod of length 2.0 m and thermal conductivity 50 W/(mK) is initially

at a uniform temperature of 100

C. One end of the rod is maintained at 200

while the other end is kept at 0

C. Calculate the rate at which heat is conducted

along the rod.

Solution

Step 1: Calculate the temperature gradient along the rod.

The temperature gradient, ∂T

∂x , is defined as the rate of change of temperature

with respect to distance. Here, it can be calculated as follows:

∂T

∂x =T2−T1

x2−x1

∂T

∂x =200C−0C

2.0m−0m

∂T

∂x = 100C/m

Step 2: Use Fourier’s Law of Heat Conduction to calculate the rate of heat

transfer.

Fourier’s Law of Heat Conduction states that the rate of heat conduction,

Q, is given by:

Q=−kA∂T

∂x

where: - kis the thermal conductivity of the material, - Ais the cross-

sectional area through which heat is being conducted.

Given that the rod is a cylinder with a cross-sectional area of A=πr2, where

ris the radius of the rod, we can calculate Qas follows:

Q=−kA∂T

∂x

Q=−50 W/(mK) ×π×(0.5m)2×100C/m

Q=−50 W/(mK) ×π×0.25m2×100C/m

Q=−50 W/(mK) ×π×0.25m2×100K

Q=−50 W/(mK) ×π×0.25 ×100

Q=−50 W/(mK) ×π×25

Q=−50 W/(mK) ×25π

Q=−1250πW≈ −3935.98 W

Therefore, the rate at which heat is conducted along the rod is approximately

3935.98 W.

Question 25

Question

A 500 g block of copper is heated to 100

C and then 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? The specific heat capacity of copper is 0.385 J/g

and for water is 4.18 J/g

Solution

Step 1: Calculate the initial heat of the copper block. The formula for heat

energy is given by Q=mc∆T, where mis the mass, cis the specific heat

capacity, and ∆Tis the change in temperature.

Given: mcopper = 500 g ccopper = 0.385 J/g

C ∆Tcopper = 100 −20 = 80

Calculating Qcopper:Qcopper = (500 g)(0.385 J/g

C)(80

C) = 15400 J

Step 2: Calculate the initial heat of the water. Given: mwater = 500 g

cwater = 4.18 J/g

C ∆Twater =Tfinal −20

Since the copper block and water reach thermal equilibrium, the total heat

lost by the copper block is equal to the total heat gained by the water. Qcopper =

Qwater

Therefore, mccopper∆Tcopper =mcwater∆Twater

Plugging in the values gives: (500 g)(0.385 J/g

C)(80

C) = (500 g)(4.18 J/g

C)(Tfinal−

20)

Solving for Tfinal: 15400 = 2090(Tfinal −20) Tfinal −20 = 15400

2090 Tfinal =

15400

2090 + 20 Tfinal ≈27.8

Therefore, the final temperature of the system will be approximately 27.8

Question 26

Question

A copper rod of length 2 m and cross-sectional area 5 cm2is initially at a tem-

perature of 100◦C. The rod is then immersed in a water bath at 0◦C. If the

thermal conductivity of copper is 390 W/mK and the heat transfer coefficient

between the rod and water is 100 W/m2K, calculate the time taken for the tem-

perature of the rod to drop to 10◦C. Assume the ambient temperature remains

constant.

Solution

Step 1: Calculate the rate of heat transfer using Newton’s Law of Cooling. The

rate of heat transfer, ˙

Q, between the rod and the water is given by Newton’s

Law of Cooling: ˙

Q=hA(Trod −Twater)

where his the heat transfer coefficient, Ais the cross-sectional area of the rod,

Trod is the temperature of the rod, and Twater is the temperature of the water.

Given that h= 100 W/m2K, A= 5 ×10−4m2,Trod = 100◦C= 373 K, and

Twater = 0◦C= 273 K, we have:

Q= 100 ×5×10−4×(373 −273) = 5 W

Step 2: Calculate the rate of temperature change. The rate of temperature

change of the rod, dTrod

dt , is given by Fourier’s Law of heat conduction:

dTrod

dt =kA

CρL (Tambient −Trod)

where kis the thermal conductivity of copper, Cis the specific heat capacity

of copper, ρis the density of copper, Lis the length of the rod, Tambient is

the ambient temperature, and Trod is the temperature of the rod. Given that

k= 390 W/mK, C= 0.39 J/gK, ρ= 8.96 g/cm3(or 8960 kg/m3), L= 2 m,

Tambient = 0◦C= 273 K, and Trod = 373 K, we have:

dTrod

dt =390 ×5×10−4

8960 ×0.39 ×2(273 −373) = −0.028 K/s

Step 3: Calculate the time taken for the temperature to drop to 10◦C. To

find the time taken for the temperature of the rod to drop from 100◦C to 10◦C,

we use the relation:

∆T=Zt

dTrod

dt dt

∆T=Z283

373

(−0.028)dt

∆T= 90 K = −0.028t

t=90

0.028 ≈3214 s

Therefore, the time taken for the temperature of the rod to drop to 10◦C is

approximately 3214 seconds.

Question 27

Question

A metallic block with a mass of 2 kg initially at a temperature of 200

C is

dropped into a container filled with 5 kg of water at room temperature of 20

If the final equilibrium temperature of the system is 30

C, calculate the specific

heat capacity of the metal. Assume no heat is lost to the surroundings.

Solution

Step 1: Calculate the heat absorbed by the metal block to raise its temperature

to the final equilibrium temperature. The heat absorbed by the metal block can

be calculated using the formula:

Qmetal =mc∆T

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

metal (in J/kg

C), ∆T= (30 −200)C=−170C(change in temperature).

Therefore,

Qmetal = 2c(−170) = −340cJ.

Step 2: Calculate the heat lost by the water to lower its temperature to the

final equilibrium temperature. The heat lost by the water can be calculated

using the formula:

Qwater =mc∆T

where: m= 5 kg (mass of water), c= 4200 J/kg

C (specific heat capacity of

water), ∆T= (30 −20)C= 10C(change in temperature).

Therefore,

Qwater = 5(4200)(10) = 210000 J.

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

metal must be equal to the heat lost by the water.

Qmetal =Qwater

−340c= 210000

c=210000

340 ≈617.65 J/kg

Therefore, the specific heat capacity of the metal is approximately 617.65 J/kg

Question 28

Question

A copper block of mass 0.5 kg is initially at a temperature of 100

C. It is then

placed into a calorimeter containing 0.2 kg of water at a temperature of 20

If the final equilibrium temperature of the system is 25

C, what is the specific

heat capacity of the copper block?

Solution

Step 1: Calculate the heat gained by the water to reach the final equilibrium

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

Qwater =mc∆T

where: m= 0.2 kg (mass of water), c(specific heat capacity of water) = 4186

J/kg

C, ∆T= 25 −20 = 5C.

So,

Qwater = 0.2×4186 ×5 = 4186 J

Step 2: Calculate the heat lost by the copper block. The heat lost by the

copper block is given by the formula:

Qcopper =mc∆T

where: m= 0.5 kg (mass of copper), c(specific heat capacity of copper), ∆T=

100 −25 = 75C.

Step 3: Since the system reaches thermal equilibrium, the heat lost by the

copper block is equal to the heat gained by the water. Thus, we have:

Qwater =Qcopper

0.2×4186 ×5=0.5×c×75

Step 4: Solve for the specific heat capacity of the copper block, c.

0.2×4186 ×5=0.5×c×75

4186 ×1 = c×75

c=4186

c= 55.8 J/kg

Therefore, the specific heat capacity of the copper block is 55.8 J/kg

Question 29

Question

A 500 g block of copper is heated from 20

C to 90

C. Calculate the amount

of heat required to raise the temperature of the copper block. (Specific heat

capacity of copper = 0.386 J/g

Solution

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

(Ti) is 20

C and the final temperature (Tf) is 90

C, the change in temperature

(∆T) can be calculated as:

∆T=Tf−Ti= 90C−20C= 70C

Step 2: Calculate the amount of heat. The amount of heat (Q) required to

raise the temperature of a substance can be calculated using the formula:

Q=mc∆T

where: - mis the mass of the substance (500 g) - cis the specific heat capacity

of the substance (0.386 J/g

C) - ∆Tis the change in temperature (70

Substitute the given values into the formula:

Q= (500 g)(0.386 J/g

C)(70C)

Step 3: Perform the calculation.

Q= 500 ×0.386 ×70 = 13510 J

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

block from 20

C to 90

C is 13510 J.

Question 30

Question

A steel rod of length Land cross-sectional area Ais initially at a temperature

T1. It is then heated until its final temperature is T2. The coefficient of linear

expansion of steel is α. Calculate the heat Qsupplied to the rod during this

process.

Solution

Step 1: Calculate the change in length of the steel rod due to the temperature

change. The change in length ∆Lof the steel rod can be calculated using the

formula for linear expansion:

∆L=αL(T2−T1)

Step 2: Calculate the change in volume of the steel rod. The change in

volume ∆Vof the steel rod is given by:

∆V=A∆L=AαL(T2−T1)

Step 3: Calculate the work done on the steel rod. The work done on the

steel rod is given by the formula:

W=P∆V

where Pis the pressure. Since the pressure is constant, we can simplify this to:

W=P∆V=P AαL(T2−T1)

Step 4: Calculate the heat supplied to the steel rod. The heat Qsupplied to

the steel rod is equal to the work done on the rod plus the increase in internal

energy:

Q=W+ ∆U

Since the process is isobaric, the increase in internal energy is given by:

∆U=nCv∆T

where nis the number of moles, Cvis the molar specific heat at constant volume,

and ∆T=T2−T1.

Step 5: Putting it all together, we find:

Q=P AαL(T2−T1) + nCv(T2−T1)

Question 31

Question

A metal sphere with a radius of 5 cm is heated until its temperature reaches

150◦C. If the initial temperature of the sphere was 20◦C, calculate the amount

of heat transferred to the sphere. (Assume the specific heat capacity of the

metal is 0.5 J/g◦C and the density of the metal is 8 g/cm3.)

Solution

Step 1: Calculate the mass of the metal sphere.

Given that the density of the metal is 8 g/cm3and the radius of the sphere is 5

cm, we can calculate the volume and mass of the sphere using the formula for

the volume of a sphere V=4

3πr3and the density formula m=ρV .

Volume of sphere = 4

3π(5 cm)3=500

3πcm3

Mass of sphere = 8 g/cm3×500

3πcm3≈4188.79 g

Step 2: Calculate the change in temperature.

The change in temperature of the sphere is:

∆T= 150◦C−20◦C = 130◦C

Step 3: Calculate the amount of heat transferred.

The amount of heat (Q) transferred to the sphere is given by the formula:

Q=mc∆T

Substitute the mass of the sphere (m= 4188.79 g), specific heat capacity of

the metal (c= 0.5 J/g◦C), and change in temperature (∆T= 130◦C) into the

formula and solve for Q.

Q= 4188.79 g ×0.5 J/g◦C×130◦C = 272252.35 J

Therefore, the amount of heat transferred to the metal sphere is 272252.35

Question 32

Question

A block of aluminum with a mass of 0.5 kg and an initial temperature of 100

C is

placed in a container of water at 20

C. If the final temperature of the aluminum

and water is 30

C, calculate the mass of water in the container. Assume no heat

is lost to the surroundings.

Solution

Step 1: Calculate the heat gained by the aluminum block. The specific heat

capacity of aluminum is 900 J/kg

C. The formula for calculating heat energy is

Q=mc∆T, where: - Qis the heat energy, - mis the mass of the substance,

-cis the specific heat capacity of the substance, and - ∆Tis the change in

temperature.

Given: - mAluminum = 0.5 kg, - cAluminum = 900 J/kg

C, - Ti,Aluminum =

100

C, and - Tf,Aluminum = 30

Plugging in the values, we get: QAluminum =mAluminum·cAluminum·∆TAluminum

QAluminum = 0.5 kg ·900 J/kg

C·(30 −100)

CQAluminum = 0.5 kg ·900 J/kg

C·

(−70)

CQAluminum =−31,500 J

Step 2: Calculate the heat lost by the water. Since there is no heat lost

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

gained by the water. The specific heat capacity of water is 4186 J/kg

Let mWater be the mass of water in kg. The initial temperature of the water

is 20

C, and the final temperature is 30

QWater =mWater ·cWater ·∆TWater QWater =mWater ·4186 J/kg

C·(30−20)

QWater =mWater ·4186 J/kg

C·10

CQWater = 41860 J/kg ·mWater

Step 3: Equate the heat gained and lost to find the mass of water. QAluminum =

QWater −31,500 J = 41860 J/kg ·mWater

Now, solve for mWater:mWater =−31,500 J

41860 J/kg mWater =−0.75 kg

However, mass cannot be negative, so we discard the negative sign. There-

fore, the mass of water in the container is 0.75 kg .

Question 33

Question

A brass rod with a length of 1.5 m is initially at a temperature of 100

C. One

end of the rod is placed in boiling water at 100

C, while the other end is placed

in an ice bath at 0

C. If the rod is a perfect conductor of heat and has a uniform

cross-sectional area of 4 ×10−4m2, calculate the steady-state temperature of

the rod.

Given: - Thermal conductivity of brass, k= 109 W/(m

K) - Heat capacity

of brass, c= 0.38 J/(g

K) - Density of brass, ρ= 8500 kg/m3

Area = 4 ×10−4m2,Length = 1.5 m,Initial temperature = 100C

Solution

Step 1: Calculate the rate of heat transfer through the rod using Fourier’s Law

of Heat Conduction:

Q=k·A·(T2−T1)

where: Q= rate of heat transfer, k= thermal conductivity of brass, A= cross-

sectional area of the rod, T2= temperature of the boiling water (100

C), T1=

initial temperature of the rod (100

C), L= length of the rod.

Plugging in the values:

Q=109 ×4×10−4×(100 −100)

1.5= 0 W

Since no heat is entering the system, we can conclude that the rod will

eventually reach a steady-state temperature.

Step 2: Calculate the steady-state temperature of the rod. In steady state,

the rate of heat transfer into the left end of the rod equals the rate of heat

transfer out of the right end of the rod.

Using the thermal equilibrium condition, we have:

m·c·(Tfinal −100) = Q

where: m=ρ·V= mass of the rod, V=A·L= volume of the rod, c= heat

capacity of brass, Tfinal = final temperature of the rod.

First, calculate the mass of the rod:

m=ρ·V=ρ·A·L= 8500 ×4×10−4×1.5=5.1 kg

Substitute the values and solve for Tfinal:

5.1·0.38 ·(Tfinal −100) = 0

1.938 ·(Tfinal −100) = 0

Tfinal = 100

Therefore, the steady-state temperature of the rod is 100

Question 34

Question

A copper rod of length 2 m has a square cross-section with sides of length 1

cm. The rod is initially at a temperature of 100 ◦C. If heat is supplied to the

rod at a rate of 50 W, how long will it take for the temperature at the center

of the rod to reach 200 ◦C? Assume the thermal conductivity of copper is 400

W/(m·K), its density is 8.96 ×103kg/m3, and its specific heat capacity is 385

J/(kg·K).

Solution

Step 1: Calculate the area of the cross-section of the copper rod. The area Aof

the square cross-section is given by

A= (side length)2= (0.01 m)2= 0.0001 m2

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

is given by

V= (area of cross-section) ×length = 0.0001 m2×2 m = 0.0002 m3

Step 3: Calculate the mass of the copper rod. The mass mof the rod is

given by

m= density ×volume = 8.96 ×103kg/m3×0.0002 m3= 1.792 kg

Step 4: Calculate the thermal energy required to heat the rod. The thermal

energy Qrequired to heat the rod from 100 ◦C to 200 ◦C is given by

Q=mc∆T

where ∆T= 200 −100 = 100 K is the temperature change, c= 385 J/(kg ·K)

is the specific heat capacity of copper, and m= 1.792 kg is the mass of the rod.

Thus,

Q= 1.792 kg ×385 J/(kg ·K) ×100 K = 69280 J

Step 5: Calculate the rate of heat transfer. The rate of heat transfer dQ

dt is

given as 50 W. We can calculate the time taken for the rod to reach 200 ◦C by

rearranging the formula for thermal energy:

dt = 50 W

⇒50 = mcdT

⇒dT

dt =50

Substitute m= 1.792 kg and c= 385 J/(kg ·K) to find

dt =50

1.792 ×385 K/s

Step 6: Calculate the time taken for the temperature at the center of the

rod to reach 200 ◦C. Given that the initial temperature is 100 ◦C and the final

temperature is 200 ◦C with a temperature change of 100 K, the time taken t

can be calculated as

t=∆T

=100

1.792×385

=100 ×1.792 ×385

50 s = 13850.24 s ≈3.847 hours

Therefore, it will take approximately 3.847 hours for the temperature at the

center of the copper rod to reach 200 ◦C when heat is supplied at a rate of 50

Question 35

Question

A sample of gas is heated at a constant volume. Initially, the gas has a pressure

of 2 atm and a temperature of 273 K. If the pressure of the gas is increased to 3

atm while maintaining the constant volume, what will be the new temperature

of the gas? Assume the gas behaves ideally.

Solution

Step 1: Use the ideal gas law to relate the initial and final conditions:

P1·V=n·R·T1

P2·V=n·R·T2

where: - P1and T1are the initial pressure and temperature of the gas, - P2and

T2are the final pressure and temperature of the gas, - Vis the volume of the

gas (constant), - nis the number of moles of gas (constant), and - Ris the ideal

gas constant.

Step 2: Since the volume and number of moles of gas are constant, we can

equate the two ideal gas law equations:

P1·T1=P2·T2

Step 3: Now substitute the given values into the equation:

2 atm ×273 K = 3 atm ×T2

546 atm ·K = 3 atm ·T2

Step 4: Solve for T2:

T2=546 atm ·K

3 atm

T2= 182 K

Therefore, the new temperature of the gas will be 182 K.