Answer:
option B
Step-by-step 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 = ?
![COP = (Q_H)/(W_(in))\\W_(in) = (Q_H)/(COP)\\W_(in) = ((0.2)/(60)* 4.18* (50-16))/(2.8)\\W_(in) = 0.169](https://img.qammunity.org/2020/formulas/engineering/college/fk51yzh45ia1l8g6wf0mw2afh86qknjz0u.png)
the required power input is 0.169 kW or 0.17 kW
hence, the correct answer is option B