1. [2082] A Carnot’s engine has 25% efficiency with a sink at 9^oC. By how many degrees should the temperature of the source be increased in order to raise the efficiency to 50%? [3] Ans: 188 K
   Solution:
   Given,
   1^st Case:
   Efficiency ([latex]\eta[/latex]) = 25% = 0.25
   Temperature of sink (T₂) = 9^oC = 9 + 273 = 282 K
   We know that,
   Or, 0.25 = 1 – [latex]\frac{282}{T_1}[/latex]
   Or, T₁ = Temperature of source = 376K
   2^nd Case:
   Let, temperature of source is to be increased by x K.
   So, [latex]\eta = 1 – \frac{T_2}{T_1}[/latex]
   
   Or, 0.5 = 1 – [latex]\frac{282}{376+x}[/latex]
   
               Or, x = 188 K
2. [2081 GIE ‘A’] For a carnot ideal engine, temperature of sink at temperature 27^oC and source is at temperature 127^oC. Calculate its efficiency. [3] Ans: 25%
   Solution:
   Temperature of sink ([latex]T_2[/latex]) = 27^oC = 27 + 273 = 300 K
   Temperature of source (T₁) = 127^oC = 127 + 273 = 400 K
   Efficiency ([latex]\eta[/latex]) =?
   We know that,
   [latex]\eta = (1-\frac{T_2}{T_1})\times 100%[/latex]
   
   = [latex](1-\frac{300}{400})\times 100%[/latex]
   
                           = 25%
3. [2081 D] A refrigerator has a coefficient of performance of 1.95. In each cycle, it absorbs [latex]3\times 10^4[/latex] J of heat from cold reservoir. How much heat is discarded to the high temperature during each cycle? [2] Ans: [latex]4.5\times 10^4[/latex] J
   Solution:
   Coefficient of performance ([latex]\beta[/latex]) = 1.95
   For refrigerator,
   Heat absorbed (Q₂) = [latex]3\times 10^4 J[/latex]
   Heat discarded (Q₁) =?
   We know that,
   [latex] \beta = \frac{\text{heat absorbed } (Q_2)}{\text{work done }(Q_1 – Q_2)}[/latex]
   
   Or, 1.95 = [latex]\frac{3\times 10^4}{Q_1 – Q_2}[/latex]
   
   Or, [latex]Q_1 – Q_2 = \frac{3\times 10^4}{1.95} [/latex]
   
   = [latex]1.54\times 10^4[/latex]
   
   Or, [latex]Q_1 = (1.54\times 10^4) + Q_2[/latex]
   
   [latex]= (1.54\times 10^4) + (3\times 10^4)[/latex]
   
   Or, [latex]Q_1 = 4.54\times 10^4 J[/latex]
4. [2080 GIE B] Two carnot engines A & B have their sources at 400 K and 350 K, and sinks at 350 K and 300 K respectively. Which engine is more efficient and by how much? [2] Ans: B is more efficient than A by 1.78%
   Solution:
   For engine A:
   Temperature of source (T₁) = 400 K
   Temperature of sink (T­₂) = 350 K
   Efficiency of A ([latex]\eta_A)[/latex]
   = [latex](1-\frac{T_2}{T_1})\times 100%[/latex]
   
           = [latex](1-\frac{350}{400})\times 100%[/latex]
   
                                       = 12.5%
   For engine B:
   Temperature of source (T₁) = 350 K
   Temperature of sink (T₂) = 300 K
   Efficiency of B ([latex]\eta_B)[/latex]
   = [latex](1-\frac{T_2}{T_1})\times 100%[/latex]
   
           = [latex](1-\frac{300}{350})\times 100%[/latex]
   
                                       = 14.28%
   Hence, engine B is more efficient than engine A by (14.28 – 12.5)% = 1.78%
5. [2079 GIE B] A carnot engine takes [latex]4.2\times 10^6[/latex] J of heat from reservoir at 627^oC and performs external work. The remaining energy is rejected into a sink at 27^oC. What is the efficiency? How much work does it perform? [1 + 1] Ans: [latex]2.8\times 10^6[/latex] J
   Solution:
   Heat taken (Q₁) = [latex]4.2\times 10^6[/latex] J
   Temperature of source (T₁) = 627^oC = 627 + 273 = 900 K
   Temperature of sink (T₂) = 27^oC = 27 + 273 = 300 K
   Efficiency ([latex]\eta[/latex]) =?
   Work done (W) =?
   We know that,
   [latex]\eta = (1 – \frac{T_2}{T_1})\times 100%[/latex]
   
               = [latex](1-\frac{300}{900})\times 100%[/latex]
   
               = 66.67%
   Again,
   [latex]\frac{Q_2}{Q_1} = \frac{T_2}{T_1}[/latex]
   
   Or, [latex]\frac{Q_2}{4.2\times 10^6} = \frac{300}{900}[/latex]
   
   Or, Q₂ = [latex]\frac{300\times (4.2\times 10^6)}{900} [/latex]
   
   = [latex]1.4\times 10^6 J[/latex]
   
   So, workdone (W) = Q₁ – Q₂
        = [latex](4.2\times 10^6) – (1.4\times 10^6)[/latex]
   
                                       = [latex]2.6\times 10^6[/latex] J
6. [2079 ‘V’] A Carnot engine working between 300 K and 600 K has a working output of 800 J per cycle. What is the amount of heat energy supplied to the engine by the source per cycle?  [3] Ans: 1600 J/cycle
   Solution:
   Temperature of sink (T₂) = 300 K
   Temperature of source (T₁) = 600 K
   Output work (W) = 800 J/cycle
   Heat supplied (Q₁) =?
   We know that,
   [latex]\frac{Q_2}{Q_1} = \frac{T_2}{T_1}[/latex]
   
   Or, [latex]\frac{Q_2}{Q_1} = \frac{300}{600} = \frac{1}{2}[/latex]
   
   Or, Q₂ = [latex]\frac{Q_1}{2}[/latex] ………………….. (i)
   Again,
   W = Q₁ – Q₂
   Or, 800 = [latex]Q_1 – \frac{Q_1}{2}[/latex]
   
   Or, 800 = [latex]\frac{Q_1}{2}[/latex]
              
   Or, Q₁ = 1600 J/cycle
7. [2076 GIE A] A carnot engine takes 10³ calories of heat from a reservoir at 227^oC and rejects heat to a reservoir at 27^oC. How much work is done by it?         Ans: 1680 J
   Solution:
   Heat taken (Q₁) = [latex]10^3 = 4.2\times 10^3[/latex] = 4200 J
   Temperature of source (T₁) = 227^oC = 227 + 273 = 500 K
   Temperature of sink (T₂) = 27^oC = 27 + 273 = 300 K
   Efficiency ([latex]\eta[/latex]) =?
   Work done (W) =?
   We know that,
   [latex]\frac{Q_2}{Q_1} = \frac{T_2}{T_1}[/latex]
   
   Or, [latex]\frac{Q_2}{4200} = \frac{300}{500}[/latex]
   
   Or, Q₂ = [latex]\frac{300\times (4200)}{500} = 2550 J[/latex]
   
   So, workdone (W) = Q₁ – Q₂
                           = [latex](4200) – (2550)[/latex]
   
                                       = 1680 J
8. [2076 GIE B] A diesel engine performs 2500 J of mechanical work and discards 4000 J of heat each cycle.

1. How much heat must be supplied to the engine each cycle?           
2. What is the thermal efficiency of the engine? Ans: 6500 J, 38.46%
   Solution:
   Work done (W) = 2500 J
   Heat discarded (Q₂) = 4000 J
   Heat supplied (Q₁) =?
   Thermal efficiency ([latex]\eta[/latex]) =?
   We know that,
   W = Q₁ – Q₂
   Or, 2500 = Q₁ – 4000
   Or, Q₁ = 2500 + 4000 = 6500 J
   Now,
   [latex]\eta = (1-\frac{Q_2}{Q_1})\times 100%[/latex]
   
               = [latex](1-\frac{4000}{6500})\times 100%[/latex]
   
                           = 38.46%

9. [2076 ‘C’] The efficiency of a Carnot cycle is 15%. If on reducing the temperature of sink by 65^oC, the efficiency becomes double, find the temperature of source and sink. Ans: 433.3 K, 368.3 K
   Solution:
   1^st Case:
   Efficiency ([latex]\eta[/latex]) = 15%
   Let, temperature of source = T₁
   Temperature of sink = T₂
   We know that,
   [latex]\eta = (1-\frac{T_2}{T_1})\times 100%[/latex]
   
   Or, 15 % = [latex](1-\frac{T_2}{T_1})\times 100%[/latex]
   
   Or, 0.15 = [latex](1-\frac{T_2}{T_1})[/latex]
   
   Or, [latex]\frac{T_2}{T_1} = 0.85[/latex]
   
   Or, [latex]T_2 = 0.85T_1[/latex] …………….. (i)
   2^nd Case:
   New temperature of sink ([latex]T_2′[/latex]) = 0.85T₁ – 65
   
   Temperature of source = T₁ (No change)
   Efficiency ([latex]\eta[/latex]) = 30% [double]
   Again,
   [latex]\eta = (1-\frac{T_2′}{T_1})\times 100%[/latex]
   
   Or, 30% = [latex](1-\frac{0.85T_1 – 65}{T_1})\times 100%[/latex]
   
   Or, 0.3 = [latex](1-\frac{0.85T_1 – 65}{T_1})[/latex]
   
   Or, [latex]\frac{0.85T_1 – 65}{T_1} = 0.7[/latex]
   
   Or, 0.85T₁ – 65 = 0.7T₁
   Or, 0.15T₁ = 65
   Or, T₁ = [latex]\frac{65}{0.15}[/latex] = 433.33 K
   
   Also, from equation (i),            
   T₂ = 0.85T₁ = [latex]0.85 \times 433.33[/latex] = 368.33 K
10. [2075 ‘A’] A carnot engine has 40% efficiency with a sink at 10^oC. By how many degrees should the temperature of the source be increased in order to raise the efficiency to 65%? Ans: 63.9^oC
    Solution:
    Given,
    1^st Case:
    Efficiency ([latex]\eta[/latex])=40% = 0.4
    Temperature of sink (T₂) = 10^oC = 10 + 273 = 283 K
    We know that,
    [latex]\eta = 1 – \frac{T_2}{T_1}[/latex]
    
    Or, 0.4 = [latex]1 – \frac{283}{T_1}[/latex]
    
    Or, T₁ = Temperature of source = 471.67K
    2^nd Case:
    Let, temperature of source is to be increased by x K.
    So, [latex]\eta = 1 – \frac{T_2}{T_1}[/latex]
    
    Or, 0.65 = [latex]1 – \frac{283}{471.67 + x}[/latex]
    
    Or, x = 336.9 K = 336.9 – 273 = 63.9^oC
11. [2074 ‘S’] A carnot engine has 50% efficiency with a sink at 9^oC. By how many degrees should temperature of source be increased in order to raise the efficiency to 70%?
    Solution:
    Given,
    1^st Case:
    Efficiency ([latex]\eta[/latex]) = 50% = 0.5
    Temperature of sink (T₂) = 9^oC = 9 + 273 = 282 K
    We know that,
    [latex]\eta = 1 – \frac{T_2}{T_1}[/latex]
    
    Or, 0.5 = [latex]1- \frac{282}{T_1}[/latex]
    
    Or, T₁ = Temperature of source = 564K
    2^nd Case:
    Let, temperature of source is to be increased by x K.
    So, [latex]\eta = 1 – \frac{T_2}{T_1}[/latex]
    
    Or, 0.7 = [latex]1 – \frac{282}{564+x}[/latex]
    
    Or, x = 376 K
12. [2074 A] An ideal heat engine operates between two reservoirs at two temperatures. In order to achieve 30% efficiency when the temperature of the sink is 50^oC, what should be the temperature of the source?
    Solution:
    Given,
    Efficiency ([latex]\eta[/latex]) = 30% = 0.3
    Temperature of sink (T₂) = 50^oC = 50 + 273 = 323 K
    We know that,
    [latex]\eta = 1 – \frac{T_2}{T_1}[/latex]
    
    Or, 0.3 = [latex]1 – \frac{323}{T_1}[/latex]
    
    Or, T₁ = Temperature of source = 461.42K = 461.42 – 273 = 188.43^oC
13. [2074 B] The source reservoir of a carnot engine is at a temperature of 400 K and takes 400 J of heat and rejects 20 J of heat to the sink reservoir in each cycle. What is the efficiency and the temperature of the sink?
    Solution:
    Given,
    Temperature of source (T₁) = 400 K
    Q₁ = 400 J
    Q₂ = 20 J
    [latex]\eta[/latex] = ?
    T₂ =?
    We know that,
    [latex]\eta = 1 – \frac{Q_2}{Q_1}[/latex]
    
    = [latex]1 – \frac{20}{400}[/latex]
    
          = 0.95 = 95%
    Then,
    [latex]\eta = 1 – \frac{T_2}{T_1}[/latex]
    
    Or, 0.95 = [latex]1 – \frac{T_2}{400}[/latex]
    
    Or, T₂ = 20 K
14. [2073 ‘C’] A carnot’s engine has 25% efficiency with a sink at 9^oC. By how many degrees should the temperature of the source be increased in order to raise the efficiency to 70%?
    Solution:
    Given,
    1^st Case:
    Efficiency ([latex]\eta[/latex]) = 25% = 0.25
    Temperature of sink (T₂) = 9^oC = 9 + 273 = 282 K
    We know that,
    [latex]\eta = 1 – \frac{T_2}{T_1}[/latex]
    
    Or, 0.25 = [latex]1- \frac{282}{T_1}[/latex]
    
    Or, T₁ = Temperature of source = 376 K
    2^nd Case:
    Let, temperature of source is to be increased by x K.
    So, [latex]\eta = 1 – \frac{T_2}{T_1}[/latex]
    
    Or, 0.7 = [latex]1 – \frac{282}{376+x}[/latex]
    
    Or, x = 564 K
15. [2072 ‘D’] A diesel engine performs 2200 J of mechanical work and discards 4300 J of heat each cycle. (i) How much heat must be supplied to the engine in each cycle? (ii) What is the thermal efficiency of the engine?
    Solution:
    Workdone (W) = 2200 J
    Heat discarded (Q₂) = 4300 J
    (i) Heat supplied (Q₁) =?
    (ii) Efficiency ([latex]\eta[/latex]) =?
    We know that,
    W = Q₁ – Q₂
    Or, Q₁ = W + Q₂
          = 2200 + 4300 = 6500 J
    For efficiency,
    [latex]\eta = 1 – \frac{Q_2}{Q_1}[/latex]
    
          = [latex]1 – \frac{4300}{6500}[/latex]
    
          = 0.3385 = 33.85%
16. [2072 ‘E’] What will be the thermal efficiency of an engine if it takes 8 KJ heat from the source and rejects 6 KJ to the sink in one cycle?
    Solution:
    Heat taken (Q₁) = 8 KJ = 8 x 1000 = 8000 J
    Heat rejected (Q₂) = 6 KJ = 6 x 1000 = 6000 J
    [latex]\eta[/latex] = ?
    We know that,
    [latex]\eta = (1 – \frac{Q_2}{Q_1})\times 100%[/latex]
    
    = [latex](1 – \frac{6000}{8000})\times 100%[/latex]
    
    = 25%