Question

A 20 watt light bulb is left burning inside a refrigerator operating on a reverse Carnot cycle. If the refrigerator also draws 20 watts of electrical power to operate its condenser, can it cool its interior below room temperature while the bulb is on? Explain why or why not

Answer 1

Step-by-step 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

`COR_(R)=(Q_(c))/(W)`

`(Q_(c))/(W)=(T_(c))/(T_(H)-T_(c))`

Put the value into the formula

`(20)/(20)=(T_(c))/(T_(H)-T_(c))`

`(T_(c))/(T_(H)-T_(c))=1`

`T_(h)=2T_(c)`

`T_(c)=(T_(h))/(2)`

`T_(c)=(27+273)/(2)`

`T_(c)=150\ K`

`T_(c)` is less than
`T_(h)`

Therefore, refrigerator cools its interior below room temperature while the bulb is on.

Hence, This is the required solution.