Question

The temperature inside a refrigerator is ${t_2}C$ and the room temperature is ${t_1}C$. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be
\[
  A.{\text{ }}\dfrac{{{t_1}}}{{{t_1} - {t_2}}} \\
  B.{\text{ }}\dfrac{{{t_1} + 273}}{{{t_1} - {t_2}}} \\
  C.{\text{ }}\dfrac{{{t_2} + 273}}{{{t_1} - {t_2}}} \\
  D.{\text{ }}\dfrac{{{t_1} + {t_2}}}{{{t_1} + 273}} \\
 \]

Answer 1

Hint- In order to deal with this question, first we will write the relation between heat and temperature for the refrigerator which is known as coefficient of performance then we will proceed further by converting temperature in Kelvin.
Formula used- \[\dfrac{{\left( {{Q_1}} \right)}}{{\left( {{Q_2}} \right)}} = \dfrac{{{T_1}}}{{{T_2}}},{Q_1} = W + {Q_2},T\left( {^0C} \right) = t + 273\left( K \right)\]

Complete step-by-step answer:

Given that:
Temperature inside refrigerator $ = {t_2}^0C$
Room temperature $ = {t_1}^0C$
For refrigerator coefficient of performance is given as
\[ = \dfrac{{{\text{Heat given to high temperature}}\left( {{Q_1}} \right)}}{{{\text{Heat taken from lower temperature}}\left( {{Q_2}} \right)}} = \dfrac{{{T_1}}}{{{T_2}}}\]
Now, we will convert temperature from Celsius to Kelvin. So we get
\[ \Rightarrow \dfrac{{{Q_1}}}{{{Q_2}}} = \dfrac{{{t_1} + 273}}{{{t_2} + 273}}.............(1)\]
As we know that room is at higher temperature and acts as a hot reservoir. And electrical energy is the input energy shown as W. So, \[{Q_1} = W + {Q_2}\]
Or we have:
\[
  {Q_1} - W = {Q_2} \\
   \Rightarrow {Q_2} = {Q_1} - W..........(2) \\
 \]
Let us substitute this equation (2) in equation (1) to get the relation:
$
  \because \dfrac{{{Q_1}}}{{{Q_2}}} = \dfrac{{{t_1} + 273}}{{{t_2} + 273}} \\
   \Rightarrow \dfrac{{{Q_1}}}{{{Q_1} - W}} = \dfrac{{{t_1} + 273}}{{{t_2} + 273}} \\
 $
Let us invert the numerator and denominator on both the side:
$
   \Rightarrow \dfrac{{{Q_1} - W}}{{{Q_1}}} = \dfrac{{{t_2} + 273}}{{{t_1} + 273}} \\
   \Rightarrow 1 - \dfrac{W}{{{Q_1}}} = \dfrac{{{t_2} + 273}}{{{t_1} + 273}} \\
   \Rightarrow \dfrac{W}{{{Q_1}}} = 1 - \dfrac{{{t_2} + 273}}{{{t_1} + 273}} \\
   \Rightarrow \dfrac{W}{{{Q_1}}} = \dfrac{{\left( {{t_1} + 273} \right) - \left( {{t_2} + 273} \right)}}{{{t_1} + 273}} \\
   \Rightarrow \dfrac{W}{{{Q_1}}} = \dfrac{{{t_1} - {t_2}}}{{{t_1} + 273}} \\
 $
To find the amount of heat delivered to the room for each joule of electrical energy (W= 1 J).
Let us substitute the value of W=1J in the above equation.
$
   \Rightarrow \dfrac{1}{{{Q_1}}} = \dfrac{{{t_1} - {t_2}}}{{{t_1} + 273}} \\
   \Rightarrow {Q_1} = \dfrac{{{t_1} + 273}}{{{t_1} - {t_2}}} \\
 $
Hence, the amount of heat delivered to the room for each joule of electrical energy consumed ideally will be $\dfrac{{{t_1} + 273}}{{{t_1} - {t_2}}}$
So, the correct answer is option B.

Note- 1J of heat means that one joule 's energy (thermal energy) was transmitted from the system to the surrounding environment, or vice versa because of the variation in temperature between the system and the surrounding area. In order to solve such problems students must remember the formula or relation between the heat and the temperature as mentioned above. Also students must remember the relation to convert Celsius into Kelvin.