Consider a room air conditioner using a Carnot cycle at maximum theoretical efficiency and operating between the temperatures of 19.7°C (indoors) and 28.7°C (outdoors). For each 1.00 J of heat flowing out of the room into the air conditioner

a) How much heat flows out of the air conditioner to the outdoors?
(in J)

b) By how much does the entropy of the room decrease?
(in J/K)

c) By how much does the entropy of the outdoor air increase?
(in J/K)

To find the answers, we can use the formula for the efficiency of a Carnot cycle:

Efficiency = 1 - (T_low / T_high)

where T_low is the temperature of the cold reservoir (indoors) and T_high is the temperature of the hot reservoir (outdoors).

a) Using the given temperatures, we can calculate the efficiency of the Carnot cycle:

Efficiency = 1 - (19.7°C / 28.7°C)

Efficiency = 1 - (292.85 K / 301.85 K)

Efficiency = 1 - 0.96811

Efficiency = 0.03189

So, the efficiency of the Carnot cycle is approximately 0.03189.

Now, since the Carnot cycle is operating at maximum theoretical efficiency, the heat flowing out of the room into the air conditioner, which is the heat rejected to the outdoors, can be calculated using the formula:

Q_out = Efficiency * Q_in

where Q_in is the heat flowing out of the room into the air conditioner, given as 1.00 J.

a) Q_out = 0.03189 * 1.00 J

Q_out = 0.03189 J

Therefore, approximately 0.03189 J of heat flows out of the air conditioner to the outdoors.

b) The decrease in entropy of the room can be calculated using the formula:

ΔS_room = Q_in / T_room

where ΔS_room is the change in entropy of the room, Q_in is the heat flowing into the air conditioner (1.00 J), and T_room is the temperature of the room (19.7°C).

ΔS_room = 1.00 J / (19.7°C + 273.15 K)

ΔS_room = 1.00 J / 292.85 K

ΔS_room = 0.003413 J/K

Therefore, the entropy of the room decreases by approximately 0.003413 J/K.

c) The increase in entropy of the outdoor air can be calculated using the formula:

ΔS_outdoor = Q_out / T_outdoor

where ΔS_outdoor is the change in entropy of the outdoor air, Q_out is the heat flowing out of the air conditioner to the outdoors (0.03189 J), and T_outdoor is the temperature of the outdoor air (28.7°C).

ΔS_outdoor = 0.03189 J / (28.7°C + 273.15 K)

ΔS_outdoor = 0.03189 J / 301.85 K

ΔS_outdoor = 0.000105718 J/K

Therefore, the entropy of the outdoor air increases by approximately 0.000105718 J/K.

To answer these questions, we can use the principles of thermodynamics and the Carnot cycle.

The Carnot cycle is a theoretical thermodynamic cycle that is considered to have the maximum possible efficiency for a heat engine operating between two temperatures. It consists of four processes: isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression.

In this case, the temperatures given are 19.7°C (292.85 K) for the indoors temperature and 28.7°C (301.85 K) for the outdoors temperature.

a) To find the amount of heat flowing out of the air conditioner to the outdoors, we need to calculate the efficiency of the Carnot cycle and then use the given 1 J of heat flowing out of the room as a reference.

The efficiency of the Carnot cycle is given by the formula:

Efficiency = 1 - (Tc/Th),

where Tc is the temperature of the cold reservoir (indoors) and Th is the temperature of the hot reservoir (outdoors).

Efficiency = 1 - (292.85 K / 301.85 K) = 1 - 0.9702 = 0.0298 (approximately)

Since the efficiency represents the ratio of usable work output to the total heat input, we can calculate the heat flowing out of the air conditioner (Qc) to the outdoors using the formula:

Qc = Efficiency * Qh,

where Qh is the heat flowing into the air conditioner (1 J).

Qc = 0.0298 * 1 J = 0.0298 J (approximately)

Therefore, approximately 0.0298 J of heat flows out of the air conditioner to the outdoors.

b) To find the decrease in entropy of the room, we need to calculate the heat transfer (Qc) and divide it by the temperature (Tc) in Kelvin.

Entropy change = Qc / Tc,

Entropy change = 0.0298 J / 292.85 K = 1.017 x 10^-4 J/K (approximately)

Therefore, the entropy of the room decreases by approximately 1.017 x 10^-4 J/K.

c) To find the increase in entropy of the outdoor air, we need to calculate the heat transfer (Qh) and divide it by the temperature (Th) in Kelvin.

Entropy change = Qh / Th,

Entropy change = 1 J / 301.85 K = 3.31 x 10^-3 J/K (approximately)

Therefore, the entropy of the outdoor air increases by approximately 3.31 x 10^-3 J/K.