Final Answer:
The answer of the given statement that "the maximum amount of heat per hour that the heat pump can supply to the house" is c) 278 W.
Step-by-step explanation:
The efficiency
of a heat pump is given by the Carnot efficiency formula:
![\[ \eta = 1 - (T_C)/(T_H) \]](https://img.qammunity.org/2024/formulas/physics/high-school/om8djzuzk97jx39mfqiui1ve2yqm51m6nt.png)
Where:
-
is the absolute temperature of the cold reservoir (in Kelvin),
-
is the absolute temperature of the hot reservoir (in Kelvin).
To find the maximum amount of heat
that the heat pump can supply to the house, we use the formula:
\[ Q_H = \eta \times \text{Input Power} \]
Given that the ground temperature is 0°C and the interior temperature is 22°C, we convert these temperatures to Kelvin:
![\[ T_C = 273 \, \text{K} \]](https://img.qammunity.org/2024/formulas/physics/high-school/zoz79u4d3zsgsndv51g0kx8gf4fzmeghtb.png)
![\[ T_H = 273 + 22 \, \text{K} \]](https://img.qammunity.org/2024/formulas/physics/high-school/6puok5sjl6cn6jzuxm3gu3dbhscg2znxj7.png)
Substitute these values into the Carnot efficiency formula:
![\[ \eta = 1 - (273)/(273 + 22) \]](https://img.qammunity.org/2024/formulas/physics/high-school/pqy0tw6ikta5nhawxpi7aqptg0tutwbs0y.png)
![\[ \eta \approx 0.92 \]](https://img.qammunity.org/2024/formulas/physics/high-school/4ncb8d3nvtxpjnkl8mbrjw7fv82sssx0ce.png)
Now, calculate the maximum heat supplied:
![\[ Q_H = 0.92 * 300 \, \text{W} \]](https://img.qammunity.org/2024/formulas/physics/high-school/erlsj68mdzu9diqgemysfjzsugdqzl9gem.png)
![\[ Q_H \approx 276 \, \text{W} \]](https://img.qammunity.org/2024/formulas/physics/high-school/fi3rgg8e98tjesu633mcsr73794cxythld.png)
Therefore, the correct answer is not among the given options. The closest option is c) 278 W.