# Questions of Advanced Thermodynamics and Answers with Explanation

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**Problem 4****:-**

A** **spherical constant temperature heat source of radius r1 is at the center of a uniform solid sphere of radius r2. The rate at which heat is transferred through the surface of the sphere is propertional to

(A) r22 – r12

(B) r2 – r1

(C) ln r1 – ln r2

(D) 1/r2 – 1/r1

(E) (1/r2 – 1/r1)-1

**Solution:-**

The correct option is (A).

The rate *H* at which heat is transferred through the slab is,

(a) directly proportional to the area (*A*) available.

(b) inversely proportional to the thickness of the slab Δ*x*.

(c) directly proportional to the temperature difference Δ*T*.

So, *H* = *kA* Δ*T*/ Δ*x*

Where *k* is the proportionality constant and is called thermal conductivity of the material.

From above we know that, the rate *H* at which heat is transferred through the slab is directly proportional to the area (*A*) available.

Area *A* of solid sphere is defined as,

*A* = 4π*r*2

Here *r* is the radius of sphere.

So, the area *A*1 of uniform small solid sphere having radius *r*1 will be,

*A*1 = 4π*r*12

And, the area *A*2 of uniform large solid sphere having radius *r*2 will be,

*A*2 = 4π*r*22

Thus the area *A* from which heat is transferred through the surface of the sphere will be the difference of area of uniform large solid sphere* A*2 and small solid sphere* A*1.

So, *A* = *A*2 - *A*1 = 4π*r*22 - 4π*r*12 = 4π (*r*22 - *r*12)

Since the rate *H* at which heat is transferred through the slab is directly proportional to the area (*A*) available, therefore the rate at which heat is transferred through the surface of the sphere is proportional to *r*22 - *r*12.

From the above observation we conclude that, option (A) is correct.

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**Problem 5****:-**

What would be the most likely value for *C*T, the molar heat capacity at constant temperature?

(A) 0

(B) 0 < *C*T < *C*v

(C) *C*v < *C*T < Cp

(D) *C*T =

**Solution:-**

The correct option is (A).

Molar heat capacity at constant temperature (*C*T) is defined as the amount of heat required to raise the temperature of 1 g of gas through 1̊ C keeping its temperature constant. But it is impossible to rise the temperature of 1 g of gas through 1̊ C keeping its temperature constant. So the two statements contradict to each other. Thus the most likely value for molar heat capacity at constant temperature (*C*T) will be zero. Therefore option (A) is correct.

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**Problem 6****:-**

Which type of ideal gas will have the largest value for Cp – Cv?

(A) Monoatomic

(B) Diatomic

(C) Polyatomic

(D) The value will be the same for all.

**Solution:-**

The correct option is (D).

The specific heat at constant volume (*C*v) for monoatomic gas is,

*C*v = 3/2 *R*

The specific heat at constant volume (*C*v) for diatomic gas is,

*C*v = 5/2 *R*

The specific heat at constant volume (*C*v) for polyatomic gas is,

*C*v = 3 *R*

But we know that, for an ideal gas,

*C*p - *C*v = *R*

So,the specific heat at constant pressure (*C*p) for monoatomic gas will be,

*C*p =* C*v + *R* = 3/2 *R* + R = 5/2 *R*

the specific heat at constant pressure (*C*p) for diatomic gas will be,

*C*p =* C*v + *R *= 5/2 *R* + R = 7/2 *R*

the specific heat at constant pressure (*C*p) for polyatomic gas will be,

*C*p =* C*v + *R *= 3 *R* + R = 4 *R*

Therefore,

*C*p - *C*v for monoatomic gas will be,

*C*p - *C*v = 5/2 *R* - 3/2 *R *= *R*

*C*p - *C*v fo diatomic gas will be,

*C*p - *C*v = 7/2 *R* - 5/2 *R* = *R*

*C*p - *C*v fo polyatomic gas will be,

*C*p - *C*v = 4 *R* - 3 *R *= *R*

From the above observation we conclude that, the value of *C*p - *C*v will be same for all ideal gas. Therefore option (D) is correct.

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**Problem 7****:-**

Consider the** **following processes that can be done on an ideal gas: constant volume, Δ*V* = 0; constant pressure, Δ*p* = 0; and constant temperature, Δ*T* = 0.

(a) For which process does W = 0?

(b) For which process does Q = 0?

(c) For which of these process does W + Q = 0?

(d) For which of these process does ΔEint = Q?

(e) For which of these process does ΔEint = W?

(A) ΔV = 0 (B) Δp = 0 (C) ΔT = 0 (D) None of these

**Solution:-**

(a) The correct option is (A) Δ*V* = 0.

The change in work done Δ*W* is defined as,

Δ*W* = *p*Δ*V*

Therefore *W* = 0 for the process where Δ*V* = 0. From the above observation we conclude that, option (A) is correct.

(b) The correct option is (D) None of these.

In adiabatic process the exchange of heat with surrounding is zero (*Q*=0). In this process, the change of volume, pressure and temperature occurs. Therefore option (D) is correct.

(c) The correct option is (C) Δ*T* = 0.

For a thermodynamic system, in which internal energy is the only type of energy the system may have, the law of conservation of energy may be expressed as,

*Q* + *W* = Δ*E*int

Here *Q* is the energy transferred (as heat) between the system and its environment, *W* is the work done on or by the system and Δ*E*int is the change in the internal energy of the system.

As *W* + *Q* = 0, it signifies that Δ*E*int = 0 and it is possible only when Δ*T* = 0. Therefor option (C) is correct.

(d) The correct option is (A) Δ*V* = 0.

For a thermodynamic system, in which internal energy is the only type of energy the system may have, the law of conservation of energy may be expressed as,

*Q* + *W* = Δ*E*int

Here *Q* is the energy transferred (as heat) between the system and its environment, *W* is the work done on or by the system and Δ*E*int is the change in the internal energy of the system.

As Δ*E*int = *Q*, it signifies that *W*=0.

The change in work done Δ*W* is defined as,

Δ*W* = *p*Δ*V*

Therefore *W* = 0 for the process where Δ*V* = 0. From the above observation we conclude that, option (A) is correct.

(e) The correct option is (D) None of these.

For a thermodynamic system, in which internal energy is the only type of energy the system may have, the law of conservation of energy may be expressed as,

*Q* + *W* = Δ*E*int

Here *Q* is the energy transferred (as heat) between the system and its environment, *W* is the work done on or by the system and Δ*E*int is the change in the internal energy of the system.

As Δ*E*int = *W*, it signifies that *Q*=0.

In adiabatic process the exchange of heat with surrounding is zero (*Q*=0). In this process, the change of volume, pressure and temperature occurs. Therefore option (D) is correct.

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**Problem 8****:-**

Consider an ideal heat pump and a perfect electric heater. The electric heater converts 100% of the electrical energy into heat energy; the heat pump converts 100% of the electrical energy into work, which then powers a Carnot refrigerator. Which is the more “efficient” way to heat a home? (Ignore maintenance or start-up costs.)

(A) The electric heater is always more efficient.

(B) The heat pump is always more efficient.

(C) The heat pump is more efficient if the outside temperature is not too cold.

(D) The heat pump is more efficient if the outside temperature is not too cold.

**Solution:-**

The correct option is (C).

A heat pump is a device that acting as a Carnot refrigerator can heat a house by transferring heat energy from the outside to the inside of the house; the process is driven by work done on the device. So, heat pump, which is also a refrigerator, is an air conditioner that can be operated in reverse to heat room. By heating the room using a perfect electric heater, there is some loss of thermal energy due to heating the coil of the electric heater.

The value of coefficient of performance *K* becomes larger as the temperature of the two reservoirs becomes more nearly the same. Thus heat pumps are more efficient in temperate climates than in climates where the outside temperature fluctuates between wide limits. From the above observation we conclude that, the heat pump is more efficient if the outside temperature is not too warm. Thus option (C) is correct.

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**Problem 9****:-**

A real engine operates at 75% of the efficiency of a Carnot engine operating betwen the same two temperatures. This engine has a power output of 100 W and discharges heat into the 27ºC low-temperature reservior at a rate of 300 J/s. what is the temperature of the high-temperature reservior?

(A) 27ºC (B) 77ºC (C) 127ºC (D) 177ºC

**Solution:-**

The correct option is (C) 127 ̊ C.

The efficiency *ε* of a Carnot engine is defined as,

*ε* = |*Q*H| -|*Q*L|/|*Q*H|

=1- |*Q*L|/|*Q*H|

= 1- *T*L/*T*H

Here *Q*L is the heat at lower temperature, *Q*H is the heat at higher temperature, *T*L is the lower temperature and *T*H is the higher temperature.

As the real engine has a power output of 100 W (100 J/s) and discharge heat into the 27 ̊ C low-temperature reservoir at a rate of 300 J/s.

Here, |*Q*H| -|*Q*L| = 100 J

And

|*Q*L| = 300 J

So, |*Q*H| -300 J = 100 J

|*Q*H| = 400 J

To obtain the efficiency *ε* of this real engine, substitute 300 J for *Q*L and 400 J for *Q*H in the equation*ε* = 1- |*Q*L/*Q*H|,

*ε* = 1- |*Q*L/*Q*H|

= 1-300 J/400 J

= 1/4

To obtain the temperature of the high-temperature region *T*H, substitute 1/4 for *ε*, 27 ̊ C for *T*L in the equation *ε* =1- *T*L/*T*H we get,

1/4=1- *T*L/*T*H

= 1-27 ̊ C/* T*H

= 1- (27+273) K/* T*H

= 1- 300 K/* T*H

So,300 K/* T*H = ¾

*T*H = 400 K

= (400 -273)̊ C

= 127 ̊ C

Therefore the temperature of the high-temperature region *T*H would be 127 ̊ C. From the above observation we conclude that, option (C) is correct.

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**Problem 10****:-**

A real has an efficiency of 33%. The engine has a work output of 24 J per cycle.

(a) How much heat energy is extracted from the high temperature reservior per cycle?

(A) 8 J (B) 16 J (C) 48 J (D) 72 J (E) The question can be answered only if the engine is a carnot engine

(b) How much heat energy is discharged into the low-temperature reservior per cycle?

(A) 8 J (B) 16 J (C) 48 J (D) 72 J (E) The question can be answered only if the engine is a carnot engine

(c) For this engine TL = 27ºC, What can be concluded about TH?

(A) TH = 450ºC (B) TH = 177ºC

(C) TH > 177ºC (D) TH < 177ºC

(E) 177ºC < TH < 450ºC

**Solution:- **

(a) The correct option is (D) 72 J.

The efficiency *ε* of a Carnot engine is defined as,

*ε* = |*Q*H| -|*Q*L|/|*Q*H|

=1- |*Q*L|/|*Q*H|

= 1- *T*L/*T*H

Here *Q*L is the heat at lower temperature, *Q*H is the heat at higher temperature, *T*L is the lower temperature and *T*H is the higher temperature.

As, *ε* = |*Q*H| -|*Q*L|/|*Q*H|

= ΔW/|*Q*H| (Since, |*Q*H| -|*Q*L| = ΔW)

To obtain the amount of heat energy is extracted from the higher temperature |*Q*H|, substitute 33% for *ε* and 24 J for ΔW in the equation *ε* = ΔW/|*Q*H|, we get,

*ε* = ΔW/|*Q*H|

33/100 = 24 J/|*Q*H|

Or, |*Q*H| = (2400/33) J

= 72.73 J

Rounding off to two significant figures, the amount of heat energy is extracted from the higher temperature |*Q*H| would be 72 J. From the above observation we conclude that, option (D) is correct.

(b) The correct option is (C) 48 J.

To obtain the heat energy is discharged into the low-temperature reservoir per cycle |*Q*L|, substitute 72 J for |*Q*H| and 33% for *ε* in the equation *ε* =1- |*Q*L|/|*Q*H|,

*ε* =1- |*Q*L|/|*Q*H|

33/100 = 1-|*Q*L|/72 J

|*Q*L|/72 J = 1-33/100

= 67/100

So,|*Q*L| = (67/100)×72 J

= 48.24 J

Rounding off to two significant figures, the heat energy is discharged into the low-temperature reservoir per cycle |*Q*L| would be 48 J. From the above observation we conclude that, option (C) is correct.

(c) The correct option is (D) *T*H<177 ̊ C.

To obtain *T*H, substitute 33% for *ε*, 27 ̊ C for *T*L in the equation *ε* =1- *T*L/*T*H we get,

33/100 =1- *T*L/*T*H

= 1-27 ̊ C/* T*H

27 ̊ C/* T*H =1-33/100

=67/100

*T*H =27 ̊ C×100/67

=40.30 ̊ C

<177 ̊ C

Therefore the value of *T*H will be <177 ̊ C. From the above observation we conclude that, option (D) is correct.

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**Problem 11****:-**

A Carnot engine discharges 3 J of heat into the low temperature reservior for every 2 J of work output.

(a) What is the efficiency of this Carnot engine?

(A) 1/3 (B) 2/5 (C) 3/5 (D) 2/3

(b) For this engine TL = 27ºC, What can be concluded about TH?

(A) TH = 627ºC (B) TH = 227ºC

(C) TH > 627ºC (D) TH < 227ºC

(E) 227ºC < TH < 627ºC

**Solution:-**

(a) The correct option is (A) 1/3.

The efficiency *ε* of a Carnot engine is defined as,

*ε* = 1- |*Q*L/*Q*H|

= 1- *T*L/*T*H

Here *Q*L is the heat at lower temperature, *Q*H is the heat at higher temperature, *T*L is the lower temperature and *T*H is the higher temperature.

To obtain the efficiency *ε* of this Carnot engine, substitute 2 J for *Q*L and 3 J for *Q*H in the equation *ε*= 1- |*Q*L/*Q*H|,

*ε* = 1- |*Q*L/*Q*H|

= 1-2 J/3 J

= 1/3

Thus the efficiency of this Carnot engine would be 1/3. From the above observation we conclude that, option (A) is correct.

(b) The correct option is (D) *T*H <227 ̊ C.

The efficiency *ε* of a Carnot engine is defined as,

*ε* = 1- |*Q*L/*Q*H|

= 1- *T*L/*T*H

*Q*L is the heat at lower temperature, *Q*H is the heat at higher temperature, *T*L is the lower temperature and *T*H is the higher temperature.

To obtain the efficiency *ε* of this Carnot engine, substitute 2 J for *Q*L and 3 J for *Q*H in the equation *ε*= 1- |*Q*L/*Q*H|,

*ε* = 1- |*Q*L/*Q*

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