Question

A Carnot air conditioner takes energy from the thermal energy of a room at 71°F and transfers it as heat to the outdoors, which is at 97°F. For each joule of electric energy required to operate the air conditioner, how many joules are removed from the room?

Answer 1

For each joule of electric energy required to operate the Carnot air conditioner, it can ideally remove 19.62 joules from the room.

Step-by-step explanation:

The question asks about the efficiency of a Carnot air conditioner that takes energy from a room at 71°F and transfers it as heat to the outdoors at 97°F. To find out how many joules are removed from the room for each joule of electric energy required to operate the air conditioner, we must understand the concept of the Coefficient of Performance (COP) of a Carnot refrigerator or air conditioner:

COP = Qc / W

Where Qc is the heat removed from the cold reservoir (room in this case) and W is the work input. Using the temperatures in Kelvin (Tc for the cold reservoir and Th for the hot reservoir), the COP for a Carnot refrigerator is given by:

COP = Tc / (Th - Tc)

First, we need to convert the given Fahrenheit temperatures to Kelvin:

Tc (room) = (71 + 459.67) × (5/9) = 294.26 K

Th (outdoors) = (97 + 459.67) × (5/9) = 309.26 K

Now, calculate the COP:

COP = 294.26 / (309.26 - 294.26) = 294.26 / 15 = 19.62

Therefore, for each joule of electric energy (W), the air conditioner can ideally remove 19.62 joules from the room (Qc).