Answer:
Step-by-step explanation:
Given that,
Hot temperature
T_H = 96°F
From Fahrenheit to kelvin
°K = (°F - 32) × 5/9 + 273
°K = (96 - 32) × 5/9 + 273
K = 64 × 5/9 + 273 = 35.56 + 273
K = 308.56 K
T_H = 308.56 K
Low temperature
T_L = 70°F
Same procedure to Levine
T_L = (70-32) × 5/9 + 273
T_L = 294.11 K
A carnot refrigerator working between a hot reservoir and at temperature T_H and a cold reservoir and at temperature T_L has a coefficient of performance K given by
K = T_L / (T_H - T_L)
K = 294.11 / (308.56 - 294.11)
K = 294.11 / 14.45
K = 20.36
Then, the coefficient of performance is the energy Q_L drawn from the cold reservoir as heat divided by work done,
So, for each joules W = 1J
K = Q_L / W
Then,
Q_L = K•W
Q_L = 20.36 × 1
Q_L = 20.36 J
Q_L ≈ 20J
So, approximately 20J of heats are removed from the room