Question

A refrigerator placed in a room at 300K has inside temperature 200K. How many calories of heat shall be delivered to the room for each \[2kcal\] of energy consumed by the refrigerator ideally?

Answer 1

Hint: Since a refrigerator is placed in a room at 300K has inside temperature 200K and the refrigerator is working as an ideal and consumed energy of \[2kcal\] then we have to find the heat delivered to the room.

Complete answer:
Here, it is given that the refrigerator is placed at 300K i.e. a higher temperature and inside temperature is 200K i.e. a lower temperature
\[\therefore {{T}_{H}}=300K\text{ }\left( Higher\text{ }temperature \right)\]
and \[{{T}_{L}}=200K\text{ }\left( lower\text{ }temperature \right)\]
Energy is equal to the work done by refrigerator
So, Energy \[w=2kcal\]
And we have to find the calories of heat to be delivered to the room
i.e. \[{{Q}_{H}}\]
As we know That the coefficient of is performance of refrigerator is given by
\[\Rightarrow CO{{P}_{R}}=\dfrac{\text{Desired output}}{\text{Required output}}=\dfrac{\text{Heat}}{\text{work done}}=\dfrac{{{Q}_{z}}}{W}\]
\[\Rightarrow CO{{P}_{R}}=\dfrac{{{T}_{L}}}{{{T}_{H}}-{{T}_{L}}}=\dfrac{200}{300-100}=\dfrac{200}{100}=2\]
But \[CO{{P}_{R}}=\dfrac{{{Q}_{L}}}{W}\] where \[{{Q}_{L}}\]is the heat extracted from the cold reservoir
\[\Rightarrow {{Q}_{L}}=CO{{P}_{R}}\times W=2\times 2\]
\[\Rightarrow 4kcal\]
Therefore, the external work done by the refrigerator is given by
\[W={{Q}_{H}}-{{Q}_{L}}\]
Where \[{{Q}_{H}}=\] heat released to the reservoir
\[\therefore {{Q}_{H}}=W+{{Q}_{Z}}=2+4=6kcal\]
Hence the correct answer is \[6kcal\] i.e. 6K calories of heat shall be delivered to the room.

Note:
Must be remembered that formula of coefficient of performance of refrigerator \[CO{{P}_{R}}=\dfrac{{{T}_{L}}}{{{T}_{H}}-{{T}_{L}}}\] and the external work done. \[W={{Q}_{H}}-{{Q}_{L}}\] to calculate the value of heat declined to the room. Be careful while calculating the value of the given data.