Problem — Carnot refrigerator (273 K)

Problem

A Carnot refrigerator maintains $T_f = 273$ K in an environment at $T_c = 300$ K. Calculate the COP and the work needed to extract $100$ kJ from the cold compartment.

Solution

Coefficient of performance. For a Carnot refrigerator:

$\beta = \frac{T_f}{T_c - T_f} = \frac{273}{300 - 273} = \frac{273}{27} \approx 10{,}1$

Work required. The COP is $\beta = Q_f/W$, so:

$W = \frac{Q_f}{\beta} = \frac{100 \ \text{kJ}}{10{,}1} \approx 9{,}9 \ \text{kJ}$

$\ev{\beta \approx 10{,}1, \qquad W \approx 9{,}9 \ \text{kJ}}$

To extract $100$ kJ from the cold compartment only about $9{,}9$ kJ of work are needed: the modest thermal gap ($27$ K) makes the process very efficient. The heat dumped into the environment is $Q_c = Q_f + W \approx 110$ kJ.

Collegamenti

Argomenti: Heat engines Concetti: Refrigerator and heat pump · Carnot cycle Competenze: Solving a thermodynamic cycle Oggetti: Refrigerator