A Carnot air conditioner takes energy from the thermal energy of a room at 70°F and transfers it as heat to the outdoors, which is at 96°F. For each joule of electric energy required to operate the air conditioner, how many joules are removed from the room? (Halliday, 05/2018, p. 606) Halliday, D., Resnick, R., Walker, J. (05/2018). Fundamentals of Physics, 11th Edition [VitalSource Bookshelf version]. Retrieved from vbk://9781119306856 Always check citation for accuracy before use.

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

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