Question

$200cal$ of heat is given to a heat engine so that it rejects $150cal$ of heat, if source temperature is $400K$, then the sink temperature is
(A) $300K$
(B) $200K$
(C) $100K$
(D) $50K$

Answer 1

Hint: In order to solve this question, we will first find the efficiency of the engine in terms of source heat and rejected heat using general formula and then will compare the efficiency of the engine in terms of source and sink temperature and thus we will solve for the sink temperature value.

Formula used:
1. Efficiency of an engine in terms of heat energy is calculated as:
\[\eta = \dfrac{{{Q_1} - {Q_2}}}{{{Q_1}}}\]
where \[{Q_1},{Q_2}\] are the heat given, heat rejected by the engine.

2. Efficiency of an engine in terms of source and sink temperature is calculated as:
\[\eta = \dfrac{{{T_1} - {T_2}}}{{{T_1}}}\]
where \[{T_1},{T_2}\] are the source temperature and sink temperature.

Complete answer:
According to the question, we have given that for an heat engine energy is given ${Q_1} = 200cal$ and heat rejected is given by ${Q_2} = 150cal$

The efficiency of the engine is given by:
\[\eta = \dfrac{{{Q_1} - {Q_2}}}{{{Q_1}}}\]

on Substituting the known values we get
\[
  \eta = \dfrac{{200 - 150}}{{200}} \\
  \eta = \dfrac{1}{4} \to (i) \\
 \]

Now, let us find the efficiency of the engine in terms of temperature where we have given that temperature of source is ${T_1} = 400K$

let sink temperature is ${T_2}$ then efficiency is given by:
\[\eta = \dfrac{{{T_1} - {T_2}}}{{{T_1}}}\]

on putting the values we get, also put value of efficiency from equation (i)
\[
  \dfrac{1}{4} = \dfrac{{400 - {T_2}}}{{400}} \\
  {T_2} = 300K \\
 \]

Hence, the correct answer is option (A) $300K$.

Note: It should be noted that no heat engine in hundred percent efficient means it’s never possible that the amount of energy given to the engine is converted into work completely, there is always a loss in energy.