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------------------------------------------------ In an air-conditioned room, the temperature is kept at [tex]6^\circ C[/tex]. The temperature outside is [tex]30^\circ C[/tex], and the air conditioner emits heat to the outside at a rate of [tex]14 \text{ kW}[/tex]. Assume that the air conditioner has a coefficient of performance that is 50% of the coefficient of performance of an ideal Carnot refrigerator.

a) At what rate does the air conditioner remove heat from the conditioned room? (Answer in kW)

b) Calculate the work required to supply the air conditioner. (Answer in kW)

c) Find the entropy change by the air conditioner in 1 hour. (Answer in kJ/K)

The air conditioner removes heat from the conditioned room at a rate of 98 kW. The supply of work required is 112 kW. The entropy change by the air conditioner in 1 hour is 0.050 kJ / K.

Explanation:

To calculate the rate at which the air conditioner removes heat from the conditioned room (Q_out), we can use the formula:

Q_out = Q_in - W

where Q_in is the rate of heat input to the air conditioner and W is the rate of work done by the air conditioner.

Given that the air conditioner emits heat to the outside at a rate of 14 kW, we can assume that the rate of heat input (Q_in) is also 14 kW.

Therefore, to find Q_out, we need to calculate the rate of work done (W).

The coefficient of performance (COP) of the air conditioner is 50% of the COP of an ideal Carnot refrigerator. The COP of a Carnot refrigerator is given by:

COP_Carnot = T_in / (T_out - T_in)

where T_in is the temperature inside the air-conditioned room and T_out is the temperature outside.

Given that the temperature inside the room is 6∘C and the temperature outside is 30∘C, we can calculate the COP_Carnot:

COP_Carnot = 6 / (30 - 6) = 0.25

Since the COP of the air conditioner is 50% of the COP_Carnot, the COP of the air conditioner is:

COP = 0.5 * 0.25 = 0.125

Now, we can calculate the rate of work done (W) using the formula:

W = Q_in / COP = 14 / 0.125 = 112 kW

Finally, we can calculate the rate at which the air conditioner removes heat from the conditioned room (Q_out) using the formula:

Q_out = Q_in - W = 14 - 112 = -98 kW

Since the rate of heat removal cannot be negative, we can conclude that the air conditioner removes heat from the conditioned room at a rate of 98 kW.

To calculate the supply of work required (W), we have already calculated it as 112 kW.

To find the entropy change (ΔS) by the air conditioner in 1 hour, we can use the formula:

ΔS = Q_in / T_in - Q_out / T_out

Given that Q_in is 14 kW, T_in is 6∘C (279 K), Q_out is -98 kW, and T_out is 30∘C (303 K), we can calculate the entropy change:

ΔS = 14 / 279 - (-98) / 303 = 0.050 kJ / K

Learn more about calculating the rate of heat removal, work supply, and entropy change in an air-conditioned room here:

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