Entropy in the Real World: Refrigerators — Question 5

Question 5

Freezing and then cooling ice require different refrigerator calculations

A reversible refrigerator rejects heat to 300 K. It first freezes .100 kg of water at 273 K, then cools the resulting ice to 253 K. Use $L_f=334000$ J/kg and $c_i=2100$ J/(kg K); neglect vessel heat and volume work.

1. Calculate minimum work for freezing at constant temperature.

2. Calculate the additional minimum work for cooling the ice and the total. Explain why one coefficient cannot be used for both stages.

Question 5 - Solution

Freezing and then cooling ice require different refrigerator calculations

(a) Method and calculation

$$Q_{c,\mathrm{freeze}}=mL_f=33400\,\mathrm J,\quad K_{R,273}=\frac{273}{300-273}.$$$$W_{\mathrm{freeze}}=33400\left(\frac{300}{273}-1\right)\approx3303.30\,\mathrm J.$$

(b) Interpretation and further calculation

$$W_{\mathrm{cool}}=mc_i\int_{253}^{273}\left(\frac{300}{T}-1\right)dT=210[300\ln(273/253)-20]\approx593.185\,\mathrm J.$$$$W_{\mathrm{total}}\approx3896.48\,\mathrm J.$$Freezing occurs at fixed cold-side temperature, but ice cooling progressively increases the temperature lift.

Physical check

The work is in addition to the heat removed, so rejected heat is their sum.