Question

A refrigerator placed in a room at $300K$ has inside temperature at $264K$ . How many calories of heat shall be delivered to the room for each $1kcal$ of energy consumed by the refrigerator ideally:
(A) $7.2kcal$
(B) $7.83kcal$
(C) $8.33kcal$
(D) $8.94kcal$

Answer 1

To the solve the given question, we will discuss or solve for the relation of heat absorbed and the energy consumed by the refrigerator ideally is the heat absorbed divided by the work done by refrigerator is equals to the room temperature is divided by the difference of inside temperature and the room temperature.

Complete step by step solution:
Let the Room Temperature be ${T_2}$ is given as $300K$ ,
And, let the temperature inside the refrigerator be ${T_1}$ is given as $264K$ .
Also, the heat absorbed and heat released by the refrigerator are ${Q_2}$ and ${Q_1}$ respectively.
And, the given energy consumed or work done by refrigerator be $W = 1kcal$
Now, the heat released by the refrigerator is the addition of heat absorbed and the energy consumed by the refrigerator.
So, ${Q_1} = {Q_2} + W$ but we don’t have the value of heat absorbed. So, we will find the ${Q_2}$ first.
Now, the relation of heat absorbed and the energy consumed is as:
$\dfrac{{{Q_2}}}{W} = \dfrac{{{T_2}}}{{{T_1} - {T_2}}}$
$\Rightarrow \dfrac{{{Q_2}}}{1} = \dfrac{{264}}{{300 - 264}}$
$\therefore {Q_2} = \dfrac{{264}}{{36}}kcal$
Now, we have the value of ${Q_2}$ , so we can use the value of ${Q_1}$ .
$\therefore {Q_1} = {Q_2} + W$
$\because {Q_1} = \dfrac{{264}}{{36}} + 1 = \dfrac{{300}}{{36}} = 8.33kcal$
So, the heat released by the refrigerator after the consumption of energy is $8.33kcal$ .

Hence, the correct option is C. $8.33kcal$

Note: Both heat and work refer to processes by which energy is transferred to or from a substance. When energy is exchanged between thermodynamic systems by thermal interaction, the transfer of energy is called heat. Work is the transfer of energy by any process other than heat.