[2082] A Carnot’s engine has 25% efficiency with a sink at 9oC. 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, 1st Case: Efficiency ([latex]\eta[/latex]) = 25% = 0.25 Temperature of sink (T2) = 9oC = 9 + 273 = 282 K We know that, Or, 0.25 = 1 – [latex]\frac{282}{T_1}[/latex] Or, T1 = Temperature of source = 376K 2nd 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
[2081 GIE ‘A’] For a carnot ideal engine, temperature of sink at temperature 27oC and source is at temperature 127oC. Calculate its efficiency. [3] Ans: 25% Solution: Temperature of sink ([latex]T_2[/latex]) = 27oC = 27 + 273 = 300 K Temperature of source (T1) = 127oC = 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%
[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]
[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%
[2079 GIE B] A carnot engine takes [latex]4.2\times 10^6[/latex] J of heat from reservoir at 627oC and performs external work. The remaining energy is rejected into a sink at 27oC. 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) = 627oC = 627 + 273 = 900 K Temperature of sink (T2) = 27oC = 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]
[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]
[2076 GIE A] A carnot engine takes 103 calories of heat from a reservoir at 227oC and rejects heat to a reservoir at 27oC. 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) = 227oC = 227 + 273 = 500 K Temperature of sink (T2) = 27oC = 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]
[2076 GIE B] A diesel engine performs 2500 J of mechanical work and discards 4000 J of heat each cycle.
How much heat must be supplied to the engine each cycle?
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%
[2076 ‘C’] The efficiency of a Carnot cycle is 15%. If on reducing the temperature of sink by 65oC, the efficiency becomes double, find the temperature of source and sink. Ans: 433.3 K, 368.3 K Solution: 1st 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]
Also, from equation (i), T2 = 0.85T1 = [latex]0.85 \times 433.33[/latex] = 368.33 K
[2075 ‘A’] A carnot engine has 40% efficiency with a sink at 10oC. By how many degrees should the temperature of the source be increased in order to raise the efficiency to 65%? Ans: 63.9oC Solution: Given, 1st Case: Efficiency ([latex]\eta[/latex])=40% = 0.4 Temperature of sink (T2) = 10oC = 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 2nd Case: Let, temperature of source is to be increased by x K. So, [latex]\eta = 1 – \frac{T_2}{T_1}[/latex]
[2074 ‘S’] A carnot engine has 50% efficiency with a sink at 9oC. By how many degrees should temperature of source be increased in order to raise the efficiency to 70%? Solution: Given, 1st Case: Efficiency ([latex]\eta[/latex]) = 50% = 0.5 Temperature of sink (T2) = 9oC = 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 2nd 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
[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 50oC, what should be the temperature of the source? Solution: Given, Efficiency ([latex]\eta[/latex]) = 30% = 0.3 Temperature of sink (T2) = 50oC = 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 = 188.43oC
[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]
[2073 ‘C’] A carnot’s engine has 25% efficiency with a sink at 9oC. By how many degrees should the temperature of the source be increased in order to raise the efficiency to 70%? Solution: Given, 1st Case: Efficiency ([latex]\eta[/latex]) = 25% = 0.25 Temperature of sink (T2) = 9oC = 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 2nd 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
[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%
[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]
