Explanation:
Given that,
Power drawn by the refrigerator = 20 watt
Power of burning light inside = 20 watt
Room temperature = 27°C
We need to calculate the temperature
Using formula of coefficient of refrigerator
[tex]COR_{R}=\dfrac{Q_{c}}{W}[/tex]
[tex]\dfrac{Q_{c}}{W}=\dfrac{T_{c}}{T_{H}-T_{c}}[/tex]
Put the value into the formula
[tex]\dfrac{20}{20}=\dfrac{T_{c}}{T_{H}-T_{c}}[/tex]
[tex]\dfrac{T_{c}}{T_{H}-T_{c}}=1[/tex]
[tex]T_{h}=2T_{c}[/tex]
[tex]T_{c}=\dfrac{T_{h}}{2}[/tex]
[tex]T_{c}=\dfrac{27+273}{2}[/tex]
[tex]T_{c}=150\ K[/tex]
[tex]T_{c}[/tex] is less than [tex]T_{h}[/tex]
Therefore, refrigerator cools its interior below room temperature while the bulb is on.
Hence, This is the required solution.
