Question

A reversible heat engine converts one- fourth of the heat input into work. When the temperature of the sink is reduced by $52K$, its efficiency is doubled. The temperature in Kelvin of the source will be_______.

Answer 1

Step 1:Variables used and given data

$\eta =$ Efficiency of heat engine

$W=$Work done by Heat Engine

$Q_1=$Heat input to the Source

$Q_2=$Heat rejected to the Sink

$T_1=$Temperature of the Source

$T_2=$Temperature of the Sink

${\eta }^{\text{'}}=$ New Efficiency

${T_2}^{\text{'}}=$New Temperature of the Sink

Given: $W=\frac{Q_1}{4}$ …$1$

When ${T_2}^{\text{'}}=T_2-52$, then ${\eta }^{\text{'}}=2\eta$

Step 2: Find the Temperature of the Source

Efficiency of reversible heat engine is given by

$\eta =\frac{W}{Q_1}=\frac{Q_1-Q_2}{Q_1}=\frac{T_1-T_2}{T_1}$

⇒ $\eta =\frac{{\displaystyle \frac{Q_1}{4}}}{Q_1}=1-\frac{T_2}{T_1}$

⇒ $\eta =\frac{1}{4}=1-\frac{T_2}{T_1}$ …$2$

⇒ $\frac{T_2}{T_1}=\frac{3}{4}$ …$3$

Now when the temperature of the sink is reduced by $52K$, its efficiency is doubled i.e.

$2\eta =1-\frac{(T_2-52)}{T_1}$

Using eq. $2$ in above eq., we get

$\frac{2}{4}=1-\frac{T_2}{T_1}+\frac{52}{T_1}$

Using eq. $3$ in the above eq., we get

$\frac{1}{2}=1-\frac{3}{4}+\frac{52}{T_1}$

⇒ $T_1=208K$

Hence, the Temperature of the Source is $208K$.