Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# The temperature inside & outside of the refrigerator is 260 K and 315 K respectively. Assuming that the refrigerator cycle is reversible, calculate the heat delivered to the surrounding for every joule of work done.

Last updated date: 20th Jun 2024
Total views: 54k
Views today: 0.54k
Verified
54k+ views
Hint Here, we are given the inside and outside temperature of a refrigerator. The inside of a refrigerator should be treated as a sink and the outside of the refrigerator should be taken as the source. We need to find the amount of heat delivered per unit work done. For that we first need to find the amount of heat extracted from the fridge per unit work.

Complete step by step solution
The reversible refrigeration cycle is the ideal reverse Carnot cycle or Carnot refrigeration cycle. A Carnot refrigerator is a device that removes heat $\left( {{Q_2}} \right)$ from the low temperature source or the inside of the refrigerator $\left( {{T_2}} \right)$ to a higher temperature $\left( {{T_1}} \right)$ sink or the outside of the refrigerator by using mechanical work (W).
Where ${Q_1}$ is the heat delivered to the surroundings
and ${Q_2}$ is the heat extracted from the refrigerator
The work done by the refrigerator, $W = {Q_2} - {Q_1}$.
Now, the coefficient of performance of a Carnot refrigerator is given by:
$\beta = \dfrac{{{Q_2}}}{W} = \dfrac{{{T_2}}}{{{T_2} - {T_1}}}$ ---- (1)
Where, ${T_2}$ is the temperature of the refrigerator, i.e. inside of the refrigerator
${T_1}$ is the temperature of the surroundings, i.e. outside of the refrigerator
In the question, we are given ${T_1}$ = 315 K and ${T_2}$ = 260 K. Putting these values in equation (1),
$\Rightarrow \dfrac{{{Q_2}}}{W} = \dfrac{{{T_2}}}{{{T_2} - {T_1}}} \\ \Rightarrow \dfrac{{{Q_2}}}{W} = \dfrac{{260}}{{315 - 260}} \\ \Rightarrow \dfrac{{{Q_2}}}{W} = \dfrac{{260}}{{55}} \\ \Rightarrow \dfrac{{{Q_2}}}{W} = \dfrac{{52}}{{11}} \\$
Now, we have to find the heat delivered to the surrounding for every joule of work done. For this we will take work done, $W = 1J$.
$\Rightarrow \dfrac{{{Q_2}}}{1} = \dfrac{{52}}{{11}} \\ \Rightarrow {Q_2} = 4.72727 \\ \Rightarrow {Q_2} \approx 4.73 \\$
This is the heat extracted from the refrigerator per joule of work done. We need to find the heat delivered to the surroundings for every 1J of work done, i.e. we need to find ${Q_1}$ for $W = 1J$.
$\because W = {Q_2} - {Q_1} \\ \Rightarrow {Q_1} = {Q_2} + W \\ \Rightarrow {Q_1} = 4.73 + 1 \\ \Rightarrow {Q_1} = 5.73 \\$

Therefore, the heat delivered to the surrounding for every joule of work done will be 5.73 J.

Note: Here is the basic diagram for the given reversible /carnot refrigerator for better understanding of the problem :