Answer:
a) Work required for air conditioner A = 354.7 J
b) Work required for air conditioner B = 310.3 J
c) The magnitude of the heat deposited outside for conditioner A = 4684.7 J
d) The magnitude of the heat deposited outside for conditioner B = 4640.3 J
Explanation:
In a carnot air conditioner, it operates like a reverse carnot engine; i.e. it removes heat from the cold reservoir (making it colder) and dumps the heat in the hot reservoir (making it hotter)
For a Carnot air conditioner,
Q꜀ is the heat removed from the colder reservoir = 4330 J for both cases
T꜀ is the temperature of the colder reservoir (temperature of the rooms) = 293 K and 296 K for both cases to be considered.
Qₕ is the heat deposited in the warmer reservoir = ? for both cases
Tₕ is the temperature of the hot reservoir (temperature of outside) = 317 K for both cases.
For Carnot air conditioners,
Qₕ = W + Q꜀ (eqn 1)
And
(Qₕ/Tₕ) - (Q꜀/T꜀) = 0 (eqn 2)
Making Qₕ the subject of formula in eqn 2
Qₕ = Tₕ (Q꜀/T꜀)
Substituting this into eqn 1
Tₕ (Q꜀/T꜀) = W + Q꜀
Q꜀ (Tₕ/T꜀) - Q꜀ = W
Q꜀ [(Tₕ - T꜀)/T꜀ ] = W
W = Q꜀ [ (Tₕ - T꜀)/T꜀ ]
For the air conditioner A,
T꜀ = 293 K, Tₕ = 317 K, Q꜀ = 4330 J, W = ?
W = Q꜀ [ (Tₕ - T꜀)/T꜀ ] = 4330 [ (317 - 293)/293] = 354.7 J
For the air conditioner B,
T꜀ = 296 K, Tₕ = 317 K, Q꜀ = 4330 J, W = ?
W = Q꜀ [ (Tₕ - T꜀)/T꜀ ] = 4330 [ (317 - 296)/296] = 310.3 J
c) Qₕ = W + Q꜀
For conditioner A,
Qₕ = 354.7 + 4330 = 4684.7 J
For conditioner B,
Qₕ = 310.3 + 4330 = 4640.3 J
