Question

An ideal refrigerator has a freezer at a temperature of \(-13\) C, The coefficient of performance of the engine is 5 . The temperature of the air to which heat is rejected will be. (A) \(325^{\circ} \mathrm{C}\) (B) \(39^{\circ} \mathrm{C}\) (C) \(325 \mathrm{~K}\) (D) \(320^{\circ} \mathrm{C}\)

Short answer

The temperature of the air to which heat is rejected will be (B) \(39^{\circ} \mathrm{C}\).

Step-by-step solution

Convert freezer temperature from Celsius to Kelvin

First, we need to convert the freezer temperature from Celsius to Kelvin using the following formula: Tc (in Kelvin) = Tc (in Celsius) + 273.15 Tc = -13 + 273.15 = 260.15 K

Re-write the formula for the COP

Now, we re-write the formula for the coefficient of performance (COP) and arrange it to solve for Th (temperature of hot reservoir): COP = (Tc) / (Th - Tc) We are given the COP as 5. Thus, the equation becomes: 5 = (260.15) / (Th - 260.15)

Solve for Th

Solve for Th: 5(Th - 260.15) = 260.15 Th - 260.15 = 52.03 Th = 312.18 K

Convert Th from Kelvin to Celsius

Finally, convert the Th from Kelvin to Celsius: Temperature of the air in Celsius = Th (in Kelvin) - 273.15 Temperature of the air in Celsius = 312.18 - 273.15 = 39.03 °C Since the question asks for the temperature of the air to which the heat is rejected, we round the result to closest integer, that is, Temperature of the air = 39 °C So the correct answer is (B) \(39^{\circ} \mathrm{C}\).