Question

A Carnot's reversible engine works with an efficiency of 50%. During each cycle, it rejects 150 calories of heat at $ 30^\circ C $ . Calculate (i) temperature of the source, (ii) work done by the engine per cycle (1 cal 4.2 J).

Answer 1

Hint : A Carnot engine operating between two given temperatures has the greatest possible efficiency of any heat engine operating between these two temperatures. All engines employing only reversible processes have this same maximum efficiency when operating between the same given temperatures. Work done by the engine is the difference between the amount of heat absorbed and the amount of heat released.

Formula used:The formulae used in the solution are given here.
The efficiency of the engine is given by $ \varepsilon = \dfrac{{{Q_1} - {Q_2}}}{{{Q_1}}} $ where $ {Q_2} $ is the amount of heat released and $ {Q_1} $ is the amount of heat absorbed.
The efficiency of the engine is given by $ \varepsilon = \dfrac{{{T_1} - {T_2}}}{{{T_1}}} $ where $ {T_1} $ is the temperature of the source and $ {T_2} $ is the sink temperature.

Complete step by step answer:
Carnot introduced the concept of absolute temperature from consideration of the reversible heat engine. A Carnot engine operating between two given temperatures has the greatest possible efficiency of any heat engine operating between these two temperatures. Furthermore, all engines employing only reversible processes have this same maximum efficiency when operating between the same given temperatures.
It has been given that the Carnot's reversible engine works with an efficiency of 50% and rejects 150 calories of heat at $ 30^\circ C $ during each cycle.
The efficiency of the engine is given by $ \varepsilon = \dfrac{{{Q_1} - {Q_2}}}{{{Q_1}}} $ where $ {Q_2} $ is the amount of heat released and $ {Q_1} $ is the amount of heat absorbed.
Given that, the heat released per cycle is 150 calories,
 $ \therefore {Q_2} = 150cal $
The efficiency $ \varepsilon = \dfrac{{50}}{{100}} = \dfrac{1}{2}. $
Thus, we can write,
 $ \dfrac{{{Q_1} - 150}}{{{Q_1}}} = \dfrac{1}{2} $
 $ \Rightarrow 2{Q_1} - {Q_1} = {Q_1} = 300cal $
 $ \therefore $ The amount of heat absorbed is $ 300cal $ .
Work done by the engine is the difference between the amount of heat absorbed and the amount of heat released.
 $ {\text{Work done = }}{Q_1} - {Q_2} $
 $ = \left( {300 - 150} \right) \times 4.2J $ , 4.2 is the constant that is multiplied when calories are converted to Joule.
 $ = 630J $
Thus, work done is $ 630J $ .
Again the efficiency of the engine is given by $ \varepsilon = \dfrac{{{T_1} - {T_2}}}{{{T_1}}} $ where $ {T_1} $ is the temperature of the source and $ {T_2} $ is the sink temperature.
 $ \dfrac{{{T_1} - {T_2}}}{{{T_1}}} = \dfrac{1}{2} $
 $ \Rightarrow 2{T_1} - {T_1} = 2{T_2} $
Given that, $ {T_2} = {30^ \circ }C = 30 + 273K = 303K $ .
Thus, the temperature of the source is,
 $ 2 \times 303 = {T_1} $
 $ \Rightarrow {T_1} = 606K $
The temperature in Celsius is, $ 606 - 273 = {333^ \circ }C. $

Note:
Carnot’s interesting result implies that 100% efficiency would be possible only if $ {T_C} = 0K $ , that is, only if the cold reservoir were at absolute zero, a practical and theoretical impossibility. But the physical implication is this—the only way to have all heat transfer go into doing work is to remove all thermal energy, and this requires a cold reservoir at absolute zero.