Question 687: Thermodynamics & KTG

Question

A Carnot engine has efficiency $\frac{1}{6}$. It becomes $\frac{1}{3}$, when the temperature of $\operatorname{sink}$ is lowered by
57 K . The temperature of the source is

Accepted Answer

We are given:
Efficiency of Carnot engine:
$ \eta = 1 - \frac{T_2}{T_1} $
where $T_1 =$ source temperature, $T_2 =$ sink temperature (both in Kelvin).
Step 1: Use first efficiency condition
$ \eta = 	frac{1}{6} = 1 - 	frac{T_2}{T_1} $
$ 	frac{T_2}{T_1} = 1 - 	frac{1}{6} = 	frac{5}{6} $
$ T_2 = 	frac{5}{6} T_1 $
Step 2: Use second efficiency condition
New efficiency $ \eta' = 	frac{1}{3}$, and new sink temp is $T_2' = T_2 - 57$.
$ 	frac{1}{3} = 1 - 	frac{T_2 - 57}{T_1} $
$ 	frac{T_2 - 57}{T_1} = 1 - 	frac{1}{3} = 	frac{2}{3} $
$ T_2 - 57 = 	frac{2}{3}T_1 $
Step 3: Solve system
From Eqn 1: $T_2 = 	frac{5}{6}T_1$.
Substitute into Eqn 2:
$ 	frac{5}{6}T_1 - 57 = 	frac{2}{3}T_1 $
$ 	frac{5}{6}T_1 - 	frac{2}{3}T_1 = 57 $
$ \left(	frac{5}{6}-	frac{4}{6}ight)T_1 = 57 $
$ 	frac{1}{6}T_1 = 57 $
$ T_1 = 342 \,	ext{K} $
Ã¢ÂÂ Answer:
The source temperature is
342 K
Correct Option: C (342 K)

Answer

171 K

Answer

399 K

Answer

342 K

Answer

285 K

Thermodynamics & KTG

A Carnot engine has efficiency \frac{1}{6}. It becomes \frac{1}{3}, when the temperature of \operatorname{sink} is lowered by 57 K . The temperature of the source is

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