Question

A refrigerator works between 4$^{\text{o}}\text{C}$ and 30$^{\text{o}}\text{C}$. It is required to remove 600 calories heat every second in order to keep the temperature of the refrigerated space constant.
The power required is (take 1cal = 4.2 joule)

Answer 1

Hint: The temperature of source is 4$^{\text{o}}\text{C}$ and the temperature of sink is 30 $^{\text{o}}\text{C}$. Calculate the rate of heat transfer with the values given in the question. After doing that you can find the work done using the formula given below. The rate of work done is nothing but power.
Formula:
$\dfrac{{{Q}_{2}}}{W}=\dfrac{{{T}_{2}}}{{{T}_{1}}-{{T}_{2}}}$
Where,
$Q$ is useful heat supplied,
W is the work done,
${{T}_{2}}$ is the temperature of sink
${{T}_{1}}$ is the temperature of source

Complete step-by-step answer:
The coefficient of performance or in short COP of a refrigerator is a ratio of useful heating or cooling provided in order to perform work.
Coefficient of performance is highly dependent on operating conditions, especially factors like absolute temperature and relative temperature between sink and source of heat. It is often graphed or averaged against the expected conditions.
$\beta \text{ = }\dfrac{{{Q}_{2}}}{W}=\dfrac{{{T}_{2}}}{{{T}_{1}}-{{T}_{2}}}$
Where,
$\beta $ is the coefficient of performance.
Given:
Useful heat = 600 cal/s = 2520 W
Work done = ?
Temperature of source = 30$^{\text{o}}\text{C}$
Temperature of sink = 4$^{\text{o}}\text{C}$
Substituting the values, we get
$\dfrac{{{\text{Q}}_{\text{2}}}}{\text{W}}\text{=}\dfrac{{{\text{T}}_{\text{2}}}}{{{\text{T}}_{\text{1}}}-{{\text{T}}_{\text{2}}}}$
$\dfrac{\text{2520}}{\text{W}}\text{=}\dfrac{277}{\text{26}}$
W = $\dfrac{2520\text{ x 26}}{277}$ = 236.5 Watt

Therefore, the correct answer is option (C).

Note: Before substituting temperature in any formula, never use the values on Celsius scale. Always use the values of temperature on Kelvin scale as the SI unit of temperature is Kelvin.