Solved Examples on Thermodynamics:-

Problem 1:-

A mixture of 1.78 kg of water and 262 g of ice at 0ºC is, in a reversible process, brought to a final equilibrium state where the water/ice ratio, by mass is 1:1 at 0ºC, (a) Calculate the entropy change of the system during this process. (b) The system is then returned to the first equilibrium state, but in an irreversible way (by using a Bunsen burner, for instance). Calculate the entropy change of the system during this process. (c) Show that your answer is consistent with the second law of thermodynamics.

Concept:-

The entropy change ΔS for a reversible isothermal process is defined as,

ΔS = Q/T

     = -mL/T

Here m is the mass, L is the latent heat and T is the temperature.

Solution:-

(a) Mass of water = 1.78 kg

Mass of ice = 262 g

So the total mass of ice and water mixture will be,

Mass of ice-water mixture = (Mass of water) + (Mass of ice)

= (1.78 kg) + (262 g) = (1.78 kg) + (262 g×10-3 kg/1 g)

= 1.78 kg + 0.262 kg = 2.04 kg

If eventually the ice and water have the same mass, then the final state will have 1.02 kg (2.04 kg/2) of each.

Thus the mass of the water that changed into ice m will be the difference of mass of water mw  and mass of final state ms.

So, m = mw - ms

To obtain mass of water that changed into ice m, substitute 1.78 kg for mass of water mw and 1.02 kg for mass of final state ms in the equation m = mw - ms,

m = mw - ms

   = 1.78 kg – 1.02 kg

  = 0.76 kg

The change of water at 0̊ C to ice at 0̊ C  is isothermal.

To obtain the change in entropy ΔS of the system during this process, substitute 0.76 kg for mass m, 333×103 J/kg for heat of fusion of water L and 273 K for T in the equation ΔS = -mL/T,

ΔS = -mL/T

     = -(0.76 kg) (333×103 J/kg )/(273 K)

     = -927 J/K

From the above observation we conclude that, the change in entropy ΔS of the system during this process will be -927 J/K.

(b) Now the system is returned to the first equilibrium state, but in an irreversible way. Thus the change in entropy ΔS of the system during this process is equal to the negative of previous case.

So, ΔS = -(- 927 J/K)

= 927 J/K

From the above observation we conclude that, the change in entropy ΔS of the system would be 927 J/K.

​(c) In accordance to second law of thermodynamics, entropy change ΔS is always  ≥  zero.

The total change in entropy will be,

ΔS = (-927 J/K) + (927 J/K) = 0

From the above observation we conclude that, our answer is consistent with the second law of thermodynamics.  

Problem 2:-

Engine A, compared to engine B, produces, per cycle, five times the work but receives three times the heat input and exhausts out twice the heat. Determine the efficiency of each engine.

Concept:-

Work done in an engine is equal to the difference of input heat and output heat.

Efficiency εA of engine A is equal to,

εA = WA/ Qi,A

Here, WA is the work done by engine A and Qi,A is the input heat of engine A. 

Efficiency εB of engine B is equal to,

εB = WB/ Qi,B

Here, WB is the work done by engine B and Qi,B is the input heat of engine B. 

Solution:-

Let work done by the engine B is WB. Since the engine A produces, per cycle, five times the work of B, so the work done WA by the engine A will be,

WA = 5 WB

Let heat input of engine B is Qi,B. Since engine A receives three times heat input of B, so the heat input Qi,A of engine A will be,

Qi,A = 3 Qi,B

Let heat output of engine B is Qo,B. Since engine A exhausts two times heat output of B, so the heat input Qo,A of engine A will be,

Qo,A = 2 Qo,B

Work done in an engine is equal to the difference of input heat and output heat.

So, WA = Qi,A - Qo,A

and

WB = Qi,B - Qo,B

Qo,B = Qi,B -WB

Substitute 5 WB for WA, 3 Qi,B for Qi,A and 2 Qo,B for Qo,A in the equation WA = Qi,A - Qo,A, we get,

WA = Qi,A - Qo,A

5 WB = 3 Qi,B - 2 Qo,B

        = 3 Qi,B – 2(Qi,B -WB)  (Since, Qo,B = Qi,B -WB)

 3 WB  = Qi,B

So, efficiency of engine B, εB will be,

εB = WB/ Qi,B

     = 1/3

And

efficiency of engine A, εA will be,

εA = WA/ Qi,A

       = 5 WB/3 Qi,B       (Since, WA = 5 WB and Qi,A = 3 Qi,B)

      = (5/3) (1/3)       (Since,WB/Qi,B = 1/3)

      = 5/9

From the above observation we conclude that, the efficiency of engine A will be 5/9 and engine B will be 1/3. 

Problem 3:-

To make some ice, a freezer extracts 185 kJ of heat at -12.0ºC. The freezer has a coefficient of performance of 5.70. The room temperature is 26.0ºC, (a) How much heat is delivered to the room? (b) How much work is required to run the freezer?

Concept:-

A freezer would like to extract as much heat QL as possible from the low-temperature reservoir (“what you want”) for the least amount of work W (“what you pay for”). So the efficiency of a freezer is defined as,

K = (what you want)/(what you pay for)

   = QL/W

and this is called coefficient of performance. The larger the value of K, the more efficient is the refrigerator.

Thus, W = QL/K

The first law of thermodynamics, applied to the working substance of the freezer, gives

W = QHQL

Here QH is the exhausted heat.

Thus exhausted heat will be,

QH = W + QL

Solution:-

(a) To obtain the heat that is delivered to the room, first we have to find out the required work W to run the freezer.

To obtain the required work W to run the freezer, substitute 185 kJ for extracted heat QL and 5.70 for the coefficient of performance K in the equation W = QL/K,

W = QL/K

   = 185 kJ/5.70

  = 32.5 kJ

To obtain the heat that is delivered to the room, substitute 32.5 kJ for work W which is required to run the freezer and 185 kJ for extracted heat QL and in the equation QH = W + QL,

QH = W + QL

     =32.5 kJ + 185 kJ

    = 217.5 kJ

Rounding off to three significant figures, the heat delivered to the room would be 218 kJ.

(b) To obtain the required work W to run the freezer, substitute 185 kJ for extracted heat QL and 5.70 for the coefficient of performance K in the equation W = QL/K,

W = QL/K

   = 185 kJ/5.70

  = 32.5 kJ

Therefore the required work W to run the freezer would be 32.5 kJ.

Problem 4:-

How much work must be done to extract 10.0 J of heat (a) from a reservior at 7ºC and transfer it to one at 27ºC by means of a refrigerator using a Carnot cycle; (b) from one at -73ºC to one at 27ºC; (c) from one at -173ºC to one at 27ºC; and (d) from one at -223ºC to one at 27ºC?

Concept:-

Coefficient of performance K of a Carnot refrigerator is defined as,

K = TL / TH - TL        …… (1)

Here TL is the lower temperature of sink and TH is the higher temperature of source.

A refrigerator would like to extract as much heat QL as possible from the low-temperature reservoir (“what you want”) for the least amount of work W (“what you pay for”). So the efficiency of a refrigerator is defined as,

K = (what you want)/(what you pay for)

   = QL/W

and this is called coefficient of performance. The larger the value of K, the more efficient is the refrigerator.

Thus, W = QL/K      …… (2)

Substitute the value of K from equation (1) in the equation W = QL/K,

W = QL/K

    = QL/( TL / TH - TL)

    = QL (TH/ TL – 1)    

Solution:-

(a) To obtain work W, substitute 10.0 J for QL, 27̊ C for TH and 7̊ C for TL in the equation W = QL (TH/ TL – 1),

W = QL (TH/ TL – 1)

    = 10.0 J (27̊ C/ 7̊ C -1)

  = 10.0 J ((27+273) K /(7+273) K -1)

 = 10.0 J (300 K/280 K – 1)

= 0.714 J

Therefore the work done would be 0.714 J.

(b) To obtain work W, substitute 10.0 J for QL, 27̊ C for TH and -73̊ C for TL in the equation W = QL (TH/ TL – 1),

W = QL (TH/ TL – 1)

    = 10.0 J (27̊ C/ (-73̊ C) -1)

  = 10.0 J ((27+273) K /(-73+273) K -1)

 = 10.0 J (300 K/200 K – 1)

= 5.00 J

Therefore the work done would be 5.00 J.

(c) To obtain work W, substitute 10.0 J for QL, 27̊ C for TH and -173̊ C for TL in the equation W = QL (TH/ TL – 1),

W = QL (TH/ TL – 1)

    = 10.0 J (27̊ C/ (-173̊ C) -1)

  = 10.0 J ((27+273) K /(-173+273) K -1)

 = 10.0 J (300 K/100 K – 1)

= 20.0 J

Therefore the work done would be 20.0 J.

d) To obtain work W, substitute 10.0 J for QL, 27̊ C for TH and -223̊ C for TL in the equation W = QL (TH/ TL – 1),

W = QL (TH/ TL – 1)

    = 10.0 J (27̊ C/ (-223̊ C) -1)

  = 10.0 J ((27+273) K /(-223+273) K -1)

 = 10.0 J (300 K/50 K – 1)

= 50.0 J

Therefore the work done would be 50.0 J.

Problem 5:-

The motor in a refrigerator has a power output of 210 W. The freezing compartment is at -3.0ºC and the outside air is at 26ºC. Assuming that the efficiency is 85%of the ideal, calculate the amount of heat that can be extracted from the freezing compartment in 15 min.

Concept:-

Coefficient of performance K of a Carnot refrigerator is defined as,

K = TL / TH - TL     

Here TL is the lower temperature of sink and TH is the higher temperature of source.

Since here the efficiency is 85% of the ideal, therefore the coefficient of performance K of the refrigerator will be,

K = 0.85 (TL / TH - TL)

A refrigerator would like to extract as much heat QL as possible from the low-temperature reservoir (“what you want”) for the least amount of work W (“what you pay for”). So the efficiency of a refrigerator is defined as,

K = (what you want)/(what you pay for)

   = QL/W

and this is called coefficient of performance. The larger the value of K, the more efficient is the refrigerator.

From the above equation K = QL/W, QL will be,

QL = (K) (W)    

Work done (W) is equal to the product of power (P) and time (t).

W = (P) (t)

Solution:-

First we have to find out the coefficient of performance K and work done W.

To obtain coefficient of performance K, substitute 270 K for TL and 299 K for TH in the equation K = 0.85 (TL / TH - TL),

K = 0.85 (TL / TH - TL)

   = 0.85 (270 K / 299 K - 270 K)

  = 7.91

To obtain work done W, substitute 210 W for power P and 15 min for time t in the equation W = (P) (t),

W = (P) (t)

   = (210 W) (15 min)

   = (210 W) (15 min) (60 s/1 min)

   = (210 W) (900 s)

   = (1.89×105 Ws) (1 J/1 Ws)

  = 1.89×105 J

To obtain the amount of heat QL that can be extracted from the freezing, substitute 7.91 for coefficient performance K and 1.89×105 J for work done W in the equation

QL = (K) (W),

QL = (K) (W)

    = (7.91) (1.89×105 J)

   = 1.50×106 J

From the above observation we conclude that, the amount of heat QL that can be extracted from the freezing compartment in 15 min would be 1.50×106 J.

Problem 6:-

Apparatus that liquefies helium is in a laboratory at 296 K. The helium in the apparatus is at 4.0 K. If 150 mJ of heat is transferred from the helium, find the minimum amount of heat delivered to the laboratory.

Concept:-

Coefficient of performance K of a Carnot refrigerator is defined as,

K = TL / TH - TL        …… (1)

Here TL is the lower temperature of sink and TH is the higher temperature of source.

A refrigerator would like to extract as much heat QL as possible from the low-temperature reservoir (“what you want”) for the least amount of work W (“what you pay for”). So the efficiency of a refrigerator is defined as,

K = (what you want)/(what you pay for)

   = QL/W

and this is called coefficient of performance. The larger the value of K, the more efficient is the refrigerator.

Thus, W = QL/K      …… (2)

Substitute the value of K from equation (1) in the equation W = QL/K,

W = QL/K

    = QL/( TL / TH - TL)

    = QL (TH/ TL – 1)       …… (3)

The first law of thermodynamics, applied to the working substance of the refrigerator, gives,

W = QHQL

Here QH is the exhausted heat.

Thus exhausted heat will be,

QH = W + QL     …… (4)

Substitute the value of W from equation (3) in the equation QH = W + QL,

QH = W + QL    

      = QL (TH/ TL – 1) + QL

     = QL (TH/ TL)       

Solution:-

To obtain the minimum amount of heat delivered to the laboratory, substitute 150 mJ for QL, 296 K for TH and 4.0 K for TL in the equation QH = QL (TH/ TL),

QH = QL (TH/ TL)

      = ((150 mJ) (10-3 J/1 mJ)) (296 K/4.0 K)

      = 11 J

From the above observation we conclude that, the minimum amount of heat delivered to the laboratory would be 11 J.

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