Animated Solution for Physics - Thermodynamics: A refrigerator consumes an average $35\text{ W}$ power to operate between temperature $-10^{\circ }\text{C}$ to $25^{\circ }\text{C}$. If there is no loss of energy, then how much average heat per second does it transfer ?

Imagine you are looking at the inner workings of a refrigerator. Its primary job is to act as a heat pump. It extracts heat from a cold space (the inside of the fridge) and dumps it into a hotter space (the room it sits in).

To do this unnatural task—moving heat from cold to hot—it requires an external energy source. This is the electrical power it consumes. The problem states that the refrigerator consumes an average power of $35\text{ W}$. Since power is the rate of doing work, we can say the work done on the system per second is $W=35\text{ J/s}$.

Before we plug any numbers into thermodynamic formulas, we must address a common trap. Temperatures given in Celsius must always be converted to the absolute Kelvin scale.

The cold reservoir temperature is $T_L=-10^{\circ }\text{C}$. Converting this to Kelvin gives us $T_L=-10+273=263\text{ K}$.

Similarly, the hot reservoir (the room) temperature is $T_H=25^{\circ }\text{C}$. Converting this yields $T_H=25+273=298\text{ K}$.

Coefficient of Performance

How do we measure the "goodness" of a refrigerator? We use the Coefficient of Performance (COP). It is defined as the desired output divided by the required input. For a refrigerator, the desired output is the heat extracted from the cold space ($Q$), and the required input is the work done ($W$).

For an ideal, reversible refrigerator (like a Carnot refrigerator operating in reverse), the COP depends entirely on the operating temperatures:

Since both expressions represent the COP, we can equate them:

Now, let's substitute our known values into this elegant equation:

Look closely at the denominator on the right side. The temperature difference $298-263$ evaluates exactly to $35$.

Notice how beautifully the numbers are set up by the examiner! The $35$ on both sides cancels out perfectly, leaving us with our final answer:

This means the refrigerator extracts $263\text{ Joules}$ of heat every second from its cold interior.

Bonus Insight: If you were asked how much heat is dumped into the room every second, you would simply use energy conservation: $Q_H=Q+W=263+35=298\text{ J/s}$.