JEE Main · 2011 · 10 Apr 2011 · Offline · Heat & Thermodynamics

A Carnot engine operating between temperatures

T_{1}

and

T_{2}

has efficiency

\frac{1}{6}

. When

T_{2}

is lowered by 62

K

, its efficiency increases to

\frac{1}{3}

. Then

T_{1}

and

T_{2}

are, respectively:

(A)372 K and 330 K
(B)330 K and 268 K
(C)310 K and 248 K
(D)372 K and 310 K✓ Correct

Answer: (D)

Solution

Carnot efficiency is

η = 1 - \frac{T _{2}}{T _{1}}

. The first condition

\frac{1}{6} = 1 - \frac{T _{2}}{T _{1}}

gives

\frac{T _{2}}{T _{1}} = \frac{5}{6}

. After lowering the sink by

62 K

\frac{1}{3} = 1 - \frac{T _{2} - 62}{T _{1}}

gives

\frac{T _{2} - 62}{T _{1}} = \frac{2}{3}

. Subtracting the two,

\frac{62}{T _{1}} = \frac{5}{6} - \frac{2}{3} = \frac{1}{6}

, so

T_{1} = 372 K

, and then

T_{2} = \frac{5}{6} \times 372 = 310 K

JEE Main · 2011 · 10 Apr 2011 · Offline · Single correct option

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