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Consider a room at 20°C that is cooled by an air conditioner with a COP of 3.2 using a power input of 2 kW, with an outside temperature of 35°C. What is the constant in the heat transfer equation (Eq. 5.15) for the heat transfer from the outside into the room?

Answer :

Final answer:

The problem involves using the air conditioner's coefficient of performance (COP), power input, and temperatures to find a constant in a heat transfer equation. We find that the air conditioner removes 6.4 kW of heat from the room. However, without additional information about the specific heat transfer coefficient for the building materials, we cannot calculate the heat transfer from outside into the room.

Explanation:

The question involves using the coefficient of performance (COP), power input, and temperatures to find a constant in a heat transfer equation. The COP of an air conditioner is given by the ratio of the heat transferred and the work done, which can be expressed as Qc/W, where Qc is the heat taken from the room and W is the work done. In this case, the COP is 3.2 and the power input W is 2 kW. By rearranging the equation, we find that Qc equals the COP times W or 3.2 x 2 kW = 6.4 kW. This amount of heat is being removed from the room by the air conditioner.

heat transfer from the outside into the room would require knowledge of other factors not provided, like the specific heat transfer coefficient for the building materials. If given, one could apply Newton's law of cooling to find this constant, which states that the rate of heat transfer dQ/dt equals to -hA(T2 - T1), where A is the area of the wall, h is the heat transfer coefficient, and T2 and T1 are outside and room temperatures respectively. However, without additional information, a specific answer cannot be given.

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