JEE MainPhysicsModerateThermodynamics

An engine has an efficiency of 0.25 when temperature of sink is
reduced by58ºC, if ts efficiency is doubled, then the temperature of the
source in °C is 29n. Find n.

Show answer and solution

Here,

\eta_{1} = 1 - \frac{T_{2}}{T_{1}}

\ 0.25 = 1 - \frac{T_{2}}{T_{1}} \Rightarrow \frac{1}{4} = 1 - \frac{T_{2}}{T_{1}}

\Rightarrow \frac{T_{2}}{T_{1}} = 1 - \frac{1}{4} = \frac{3}{4}

.....(i)
According to question,

\eta_{2} = 2\eta_{1},

and

T_{2} = T_{2} - 58^{\circ}C

\therefore 2 \times \frac{1}{4} = 1 - \frac{\left( T_{2} - 58^{\circ}C \right)}{T_{1}}

\Rightarrow 1 - \frac{1}{2} = \frac{T_{2} - 58^{\circ}C}{T_{1}}

\frac{1}{2} = \frac{T_{2}}{T_{1}} - \frac{58^{\circ}}{T_{1}} \Rightarrow \frac{3}{4} - \frac{1}{2} = \frac{58}{T_{1}}

\Rightarrow T_{1} = 232^{\circ}C

Source: Grass-G / P-08 Physics paper 23 October.docx

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