Answer:
option B
Explanation:
given,
heating tap water from 16° C to 50° C
at the average rate of 0.2 kg/min
the COP of this heat pump is 2.8
power output = ?
[tex]COP = \dfrac{Q_H}{W_{in}}\\W_{in} = \dfrac{Q_H}{COP}\\W_{in} = \dfrac{\dfrac{0.2}{60}\times 4.18\times (50-16)}{2.8}\\W_{in} = 0.169[/tex]
the required power input is 0.169 kW or 0.17 kW
hence, the correct answer is option B
