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

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.