Carnot engine air conditioner

[noname1]

A Carnot air conditioner takes energy from the thermal energy of a room at 76°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?

Attempt at problem

76 + 273 = 349
96 + 273 = 369

COP = energy removed / elec energy required

COP = 349 / (369 - 349) = 17.45

17.45 = energy removed / 1 Joule

the system is always saying that my answer is incorrect but i can't figure out what

 

[vela]

Your temperatures are in Fahrenheit, not Celsius, so you're converting them to absolute temperatures incorrectly. The rest looks okay, though.

 

[noname1]

shoot i didnt notice that but anyways converted now but the system still says i got the wrong answer


(5/9)(76-32) = 297.44
(5/9)(96-32) = 308.56

COP = energy removed / elec energy required

COP = 308.56 / (308.56 - 297.44) = 27.75

27.75 = energy removed / 1 Joule

i can't find out what's wrong

 

[vela]

You're using the wrong temperature in the numerator.

 

[rdl]

i'm doing wrong

It is possible that your calculation for the coefficient of performance (COP) is incorrect. The COP is defined as the ratio of energy removed from the room to the energy required to operate the air conditioner. In this case, the energy removed from the room is the difference between the initial and final temperatures (369-349 = 20) and the energy required to operate the air conditioner is 1 joule. Therefore, the COP should be 20/1 = 20. This means that for every joule of electric energy required to operate the air conditioner, 20 joules of thermal energy are removed from the room.