A carnot engine has efficiency 1/5. Efficiency becomes 1/3when temperature of sin k is decreased by 50 K. What is the temperature of source?a)325 Kb)375 Kc)300 Kd)350 KCorrect answer is option 'B'. Can you explain this answer?

Question

Answer

Let T be temperature of source, T₁ be initial temperature of sink in the first case and T₂ be temperature of sink in second case.

For the first case,

T₁ =

For the second case,

T₁ - T₂ = 50

= 50
T = 25

×

15 = 375 K

Answer

Given: Efficiency of Carnot engine, η1 = 1/5, Efficiency after decreasing the temperature of source, η2 = 1/3 and temperature difference, ΔT = 50 K

To find: Temperature of source, T1

Concept:
The efficiency of the Carnot engine is given by η = 1 – T2/T1, where T1 is the temperature of the source, and T2 is the temperature of the sink.

Calculation:
Let us assume the temperature of the sink is T2.

Efficiency of Carnot engine, η1 = 1/5

η1 = 1 – T2/T1
1/5 = 1 – T2/T1
T2/T1 = 4/5

Now, when the temperature of source is decreased by 50 K, efficiency becomes 1/3

η2 = 1 – (T2/(T1 – 50))
1/3 = 1 – (T2/(T1 – 50))

(T2/(T1 – 50)) = 2/3
(T2/T1) * (T1/(T1 – 50)) = 2/3
(4/5) * (T1/(T1 – 50)) = 2/3
T1/(T1 – 50) = (2/3) * (5/4)
T1/(T1 – 50) = 5/6
6T1 = 5(T1 – 50)
T1 = 375 K

Therefore, the temperature of the source is 375 K.

Answer: Option B.

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