A Carnot reversible engine converts 1/6 of heat input into work. When the temperature of the sink is reduced by 62K , the efficiency of Carnot's cycle becomes 1/3 . What is the sum of the temperature (in kelvin) of the source and the sink?

Question

A Carnot reversible engine converts 1/6 of heat input into work. When the temperature of the sink is reduced by 62K , the efficiency of Carnot's cycle becomes 1/3 . What is the sum of the temperature (in kelvin) of the source and the sink? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

The efficiency of heat engine is given by

η = \frac{W}{Q} = 1 - \frac{Q _{2}}{Q _{1}} = 1 - \frac{T _{2}}{T _{1}}

where

T_{1}

is temperature of source and

T_{2}

is temperature of sink.
Given,

η_{1} = \frac{1}{6}, η_{2} = \frac{1}{3}

∴ \frac{1}{6} = \frac{T _{1} - T _{2}}{T _{1}} ... (1)

and

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

Solving Eqs.

(1)

and

(2)

, we get

T_{1} = 372 K

and

T_{2} = 310 K

T_{1} + T_{2} = 310 + 372 = 682

Q. A Carnot reversible engine converts 1/6 of heat input into work. When the temperature of the sink is reduced by 62K , the efficiency of Carnot's cycle becomes 1/3 . What is the sum of the temperature (in kelvin) of the source and the sink?

Q. A Carnot reversible engine converts $1/6$ of heat input into work. When the temperature of the sink is reduced by $62 K$ , the efficiency of Carnot's cycle becomes $1/3$ . What is the sum of the temperature (in kelvin) of the source and the sink?