A refrigerator consumes an average 35W power to operate between temperature -10°C to 25°C. If there is no loss of energy then how much average heat per second does it transfer?

Solution

Physics sir · 4 steps

Treat it as an ideal Carnot refrigerator: its COP times the work input gives the heat pulled out each second.

1. Kelvin first. Cold side: T_L = -10 + 273 = 263 K. Hot side: T_H = 25 + 273 = 298 K. The gap is T_H - T_L = 35 K.
2. Ideal COP. With no energy loss, the refrigerator works like a Carnot refrigerator. Its coefficient of performance is the heat taken from the cold side divided by the work input, and it depends only on the two temperatures.

   β = Q_L/W = T_L/(T_H - T_L) = 263/35

3. Heat per second. The work per second is W = 35 J/s, so Q_L = β W = 263/35 × 35 = 263 J/s.
4. So. The refrigerator removes 263 J/s from the cold space, which is option (a).

The idea to keep. For an ideal refrigerator, Q_L/W = T_L/(T_H - T_L) with temperatures in kelvin. The heat removed from the cold side is Q_L. The room receives Q_L + W.

Answer from the NTA final answer key. Solution by PrepWiser.