Thermodynamics
- A Carnot engine whose sink is at 300 K has an efficiency of 40%. By how much should the temperature of source be increased so as to increase, its efficiency by 50% of original efficiency ?
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We know that efficiency of Carnot Engine
= T1 - T2 T1
where, T1 is temp. of source & T2 is temp. of sink∴ 0.40 = T1 - 300 ⇒ T1 - 300 = 0.40T1 T1 0.6T1 = 300 ⇒ T1 = 300 = 3000 = 500 K 0.6 6
Now efficiency to be increased by 50%∴ 0.60 = T1 - 300 ⇒ T1 - 300 = 0.6T1 T1 0.4T1 = 300 ⇒ T1 = 300 = 300 × 10 = 750 K 0.4 4
Increase in temp = 750 – 500 = 250 KCorrect Option: B
We know that efficiency of Carnot Engine
= T1 - T2 T1
where, T1 is temp. of source & T2 is temp. of sink∴ 0.40 = T1 - 300 ⇒ T1 - 300 = 0.40T1 T1 0.6T1 = 300 ⇒ T1 = 300 = 3000 = 500 K 0.6 6
Now efficiency to be increased by 50%∴ 0.60 = T1 - 300 ⇒ T1 - 300 = 0.6T1 T1 0.4T1 = 300 ⇒ T1 = 300 = 300 × 10 = 750 K 0.4 4
Increase in temp = 750 – 500 = 250 K
- An engine has an efficiency of 1/6. When the temperature of sink is reduced by 62°C, its efficiency is doubled. Temperature of the source is
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Since efficiency of engine is η = 1 - T2 T1
According to problem,1 = 1 - T2 ......(1) 6 T1
When the temperature of the sink is reduced by 62°C, its efficiency is doubled2 
1 
= 1 - T2 ...............(2) 6 T1
Solving (1) and (2) T2 = 372 K
T1 = 99°C = Temperature of source.Correct Option: C
Since efficiency of engine is η = 1 - T2 T1
According to problem,1 = 1 - T2 ......(1) 6 T1
When the temperature of the sink is reduced by 62°C, its efficiency is doubled2 1 = 1 - T2 ...............(2) 6 T1
Solving (1) and (2) T2 = 372 K
T1 = 99°C = Temperature of source.
- When 1 kg of ice at 0°C melts to water at 0°C, the resulting change in its entropy, taking latent heat of ice to be 80 cal/°C, is
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Change in entropy is given by
dS = dQ or ∆S = ∆Q = mLT T T 273 ∆S = 1000 × 80 = 293 cal /K 273 Correct Option: D
Change in entropy is given by
dS = dQ or ∆S = ∆Q = mLT T T 273 ∆S = 1000 × 80 = 293 cal /K 273
- Two Carnot engines A and B are operated in series. The engine A receives heat from the source at temperature T1 and rejects the heat to the sink at temperature T. The second engine B receives the heat at temperature T and rejects to its sink at temperature T2. For what value of T the efficiencies of the two engines are equal?
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Efficiency of engine A, η1 = 1 - T T1 Efficiency of engine B , η2 = 1 - T2 T
Here, η1 = η2∴ T = T2 ⇒ T = √T1 T2 T1 T
Correct Option: D
Efficiency of engine A, η1 = 1 - T T1 Efficiency of engine B , η2 = 1 - T2 T
Here, η1 = η2∴ T = T2 ⇒ T = √T1 T2 T1 T
- An ideal carnot engine, whose efficiency is 40% receives heat at 500 K. If its efficiency is 50%, then the intake temperature for the same exhaust temperature is
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Efficiency of carnot engine (η1) = 40% = 0.4;
Initial intake temperature (T1) = 500K and new efficiency (η2) = 50% = 0.5. Efficiency , η = 1 - T2 or T2 = 1 - η T1 T1 Therefore in first case, T2 = 1 - 0.4 = 0.6 500
⇒ T2 = 0.6 × 500 = 300K And in second case , 300 = 1 - 0.5 = 0.5 T1 ⇒ T1 = 300 = 600 K 0.5
Correct Option: A
Efficiency of carnot engine (η1) = 40% = 0.4;
Initial intake temperature (T1) = 500K and new efficiency (η2) = 50% = 0.5. Efficiency , η = 1 - T2 or T2 = 1 - η T1 T1 Therefore in first case, T2 = 1 - 0.4 = 0.6 500
⇒ T2 = 0.6 × 500 = 300K And in second case , 300 = 1 - 0.5 = 0.5 T1 ⇒ T1 = 300 = 600 K 0.5
