Question 760: Thermodynamics & KTG

Question

A Carnot's cycle operating between $T_H=600 \mathrm{~K}$ and $T_c=300 \mathrm{~K}$ produces 1.5 kJ of mechanical work per cycle. The heat transferred to the engine by the reservoir is

Accepted Answer

To find the heat transferred to the engine, we start by determining the efficiency of a Carnot engine operating between two temperatures, $ T_H = 600 \, \mathrm{K} $ and $ T_C = 300 \, \mathrm{K} $.
The efficiency $\eta$ of a Carnot engine is given by:
$ \eta = 1 - \frac{T_C}{T_H} = 1 - \frac{300}{600} $
This simplifies to:
$ \eta = \frac{1}{2} $
Efficiency is also defined as the ratio of the work done per cycle ($W$) to the heat transferred to the engine per cycle ($Q$):
$ \eta = \frac{W}{Q} $
Given that the work done per cycle is $ 1.5 \, \mathrm{kJ} $:
$ \frac{1}{2} = \frac{1.5}{Q} $
Solving for $Q$:
$ Q = 1.5 	imes 2 = 3 \, \mathrm{kJ} $
Thus, the heat transferred to the engine per cycle is $ 3 \, \mathrm{kJ} $.

Answer

2.5 kJ

Answer

3.0 kJ

Answer

3.5 kJ

Answer

4.0 kJ

Thermodynamics & KTG

A Carnot's cycle operating between T_H=600 \mathrm{~K} and T_c=300 \mathrm{~K} produces 1.5 kJ of mechanical work per cycle. The heat transferred to the engine by the reservoir is

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