Question

The efficiency of a heat engine is $30\%$. If it gives $30\;{\rm{kJ}}$ heat to the heat sink, then it should have absorbed __________kJ heat from the heat source.
A. 9
B. 39
C. 29
D. 42.8

Answer 1

The above problem can be resolved by using the concepts and the fundamentals of the heat engine. The heat engine is one of the special machines that work on the principle of thermodynamics laws. In the heat engine analysis, when some amount of thermal energy is provided to the system, then the system will be observed to do some work. Then, the heat engine's efficiency is given by taking the ratio of the thermal energy given to the engine tank and the output obtained in the form of work.

Complete step by step answer:
Given:
The efficiency of a heat engine is, $\eta = 30\%$.
The heat output is, ${H_O} = 30\;{\rm{kJ}}$.
The expression for the efficiency of heat engine is given as,
$\eta = 1 - \dfrac{{{H_O}}}{{{H_i}}}$
Here, ${H_i}$ represents the heat absorbed from the heat source.
On solving by substituting the values in the above equation as,

\eta = 1 - \dfrac{{{H_i}}}{{{H_O}}}\\\ 0.3 = 1 - \dfrac{{{H_i}}}{{30\;{\rm{kJ}}}}\\\ {H_i} = 42.8\;{\rm{kJ}} \end{array}$$ Therefore, the heat absorbed from the heat source is of $$42.8\;{\rm{kJ}}$$ and option (D) is correct. **Note:** To resolve the above problem, one must understand the concept and the heat engine; the heat engine is that kind of machine that takes some energy and converts that fraction of energy into the mechanical work. And rest part of the energy is stored within the system in the form of internal energy. Therefore, the working principle of a heat engine is generally based on the first law of thermodynamics.