An air conditioner cools a house at [tex]T_L = 20^\circ \text{C}[/tex] with a maximum of 1.2 kW power input. The house gains 0.6 kW per degree temperature difference to the ambient, and the refrigeration COP is [tex]\beta = 0.6 \beta_{\text{Carnot}}[/tex]. Find the maximum outside temperature, [tex]T_H[/tex], for which the air conditioner provides sufficient cooling.

To find the maximum outside temperature for which the air conditioner provides sufficient cooling, we need to consider the heat transfer in the system. The rate of heat transfer from the house to the surroundings is equal to the rate of power input to the air conditioner. The formula for the coefficient of performance (COP) of a Carnot refrigerator can be used to solve for the maximum outside temperature.

Explanation:

To find the maximum outside temperature for which the air conditioner provides sufficient cooling, we need to consider the heat transfer in the system. The rate of heat transfer from the house to the surroundings is equal to the rate of power input to the air conditioner. The rate of heat transfer can be calculated using the formula: ΔQ = m * c * ΔT, where m is the mass of air, c is the specific heat capacity of air, and ΔT is the temperature difference.

The air conditioner can cool the house by 20 degrees C, so the temperature difference is 20 degrees C. The rate of heat transfer is 1.2 kW. The house gains 0.6 kW per degree temperature difference. Therefore, the rate of heat transfer from the house to the surroundings is 0.6 * 20 = 12 kW. Setting this equal to the rate of power input, we can solve for the maximum outside temperature. Using the formula for the coefficient of performance (COP) of a Carnot refrigerator, we can solve for the maximum outside temperature.

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