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# The temperature of food material in the refrigerator is $4^\circ C$ and temperature of the environment is $15^\circ C$. If a Carnot cycle is used in its working gas, then find its Carnot efficiency.(A) $0.038$(B) $0.028$(C) $0.053$ (D) $0.072$

Last updated date: 20th Jun 2024
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Hint: The Carnot efficiency can be calculated by using the equation: $\eta = 1 - \dfrac{{{T_C}}}{{{T_H}}}$. As the temperatures of food inside the refrigerator and the temperature of the environment are given, mere substitution gives the solution.

Formula used: The Carnot efficiency,
$\eta = 1 - \dfrac{{{T_C}}}{{{T_H}}}$
where ${T_C}$ is the temperature of the cold reservoir and ${T_H}$ is the temperature of the hot reservoir, both represented in absolute Kelvin.
The temperature given in Celsius scale can be converted into the Kelvin scale as follows:
${T_{Kelvin}} = {T_{Celcius}} + 273$

It is also given the temperature of the food material inside the refrigerator, that is the temperature of the cold reservoir, ${T_C} = 4^\circ C = 273 + 4 = 277K$.
${T_H} = 15^\circ C = 15 + 273 = 288K$
$\eta = 1 - \dfrac{{{T_C}}}{{{T_H}}} = 1 - \dfrac{{277}}{{288}}$
$\therefore \eta = 1 - 0.9618 = 0.0381 \approx 0.038$
So, the Carnot efficiency of the refrigerator is $0.038$ and the correct answer is option A.