A Carnot engine is operating between the two reservoirs of temperature $227 \ °C$ and $127 \ °C$ respectively. If the engine performs $1.04 \ \times \ 10^5 \ J$ work per cycle. Then, the heat absorbed from hot reservoir per cycle will be :

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A Carnot engine is operating between the two reservoirs of temperature $227 \ °C$ and $127 \ °C$ respectively. If the engine performs $1.04 \ \times \ 10^5 \ J$ work per cycle. Then, the heat absorbed from hot reservoir per cycle will be :

$2.08 \ \times \ 10^4 \ J$

$1.04 \ \times \ 10^5 \ J$

$5.20 \ \times \ 10^5 \ J$

$2.36 \ \times \ 10^5 \ J$

The correct answer is

$5.20 \ \times \ 10^5 \ J$

Carnot Engine Calculations

This problem involves calculating the heat absorbed from the hot reservoir ($Q_H$) for a Carnot engine, given the work done per cycle ($W$) and the temperatures of the hot ($T_H$) and cold ($T_C$) reservoirs.

Temperature Conversion

First, convert the given temperatures from Celsius to Kelvin (K):

Hot reservoir temperature: $T_H = 227 \ °C + 273 = 500 \ K$
Cold reservoir temperature: $T_C = 127 \ °C + 273 = 400 \ K$

Carnot Efficiency Calculation

The efficiency ($\eta$) of a Carnot engine depends only on the reservoir temperatures:

$ \eta = 1 - \frac{T_C}{T_H} $

Substitute the Kelvin temperatures:

$ \eta = 1 - \frac{400 \ K}{500 \ K} = 1 - \frac{4}{5} = 1 - 0.8 = 0.2 $

Heat Absorbed Calculation

The efficiency of any heat engine is also defined as the ratio of work done ($W$) to the heat absorbed from the hot reservoir ($Q_H$):

$ \eta = \frac{W}{Q_H} $

Rearrange the formula to solve for $Q_H$:

$ Q_H = \frac{W}{\eta} $

Given $W = 1.04 \times 10^5 \ J$ and calculated $\eta = 0.2$:

$ Q_H = \frac{1.04 \times 10^5 \ J}{0.2} $

$ Q_H = 5.20 \times 10^5 \ J $

Conclusion

The heat absorbed from the hot reservoir per cycle is $5.20 \times 10^5 \ J$.

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