Question

The motor that drives a refrigerator produces 148 W of useful power. The hot and cold temperatures of the heat reservoirs are 20.0°C and −5.0°C. What is the maximum possible amount of ice it can produce in 2.0 h from water that is initially at 8.0°C?

 

 

Answer 1

Power P = 148 W

Temperature of hot reservoir TH = 273.15 + 20 ⇒ 293.15 K

Temperature of the cold reservoir TC = 273.15 – 5°C ⇒ 268.15 K

Change in temperature in converting from water to ice ΔT = 8°C

Time taken in converting water to ice Δt = 2hrs ⇒ 2 × 3600s = 7200 s

 

The efficiency is given by

e = 1 – TC/TH

    = 1 – 268.15 K/293.15 K

   = 0.0853

 

The amount of heat rejected from water to make it to ice is

QC = QH – Wnet

      = Wnet/e – Wnet

      = Wnet(1/e – 1) (∴Wnet = P×Δt)

      = P × Δt(1/e – 1)

      = (148 W)(7200s)(1/0.0853 – 1)

      = 1.1427 × 107 J

 

The amount of heat removed from water is

QC = mcΔT + mLf

  m = QC/(cΔT + Lf)

      = 1.1427 × 107 J/[4186 J/kg.K)(8)+(333.7 × 103 J/kg)]

      = 31.1 kg

e = 0.0853

QC = 1.1427 × 107 J