A heat pump for a home has a hot reservoir temperature of 45.0°C and a cold reservoir temperature of -15.0°C. What is the best possible coefficient of performance for this heat pump?
A1.89
B3.30
C4.30
D5.30
Answer & Solution
Correct answer: D. 5.30
1. Convert both temperatures to kelvins first: Th = 45.0°C + 273 = 318 K, and Tc = -15.0°C + 273 = 258 K.
2. Find the Carnot efficiency for these reservoirs: EffC = 1 - Tc/Th = 1 - (258 K)/(318 K).
3. Dividing gives 258/318 = 0.811, so EffC = 1 - 0.811 = 0.189 (to three significant figures, 0.1887).
4. The best possible coefficient of performance is the reciprocal of this efficiency: COPhp = 1/EffC = 1/0.1887.
5. Carrying out the division: 1/0.1887 = 5.30.
6. Option A, 1.89, mistakes the Carnot efficiency itself, expressed as a raw fraction, for its reciprocal.
7. Options B and C, 3.30 and 4.30, do not follow from correctly taking the reciprocal of 0.1887, so neither matches the required calculation.
_Source: OpenStax College Physics (CC BY 4.0), Ch 15 "Thermodynamics", section 15.5 Applications of Thermodynamics: Heat Pumps and Refrigerators_
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