In one cycle, a freezer uses 585 j of electrical energy in order to remove 1750 j of heat from its freezer compartment at 10.0 f. a. what is the coefficient of performance of this freezer? b. how much heat does it expel into the room during this cycle?

(a) 0.525

(b) 1750J

Explanation:

Coefficient of Performance (COP) = Q2/(Q1 - Q2)

Q1 = heat rejected to the surrounding = 1750J

Q2 = heat absorbed in the environment = 585K

COP = 585/(1700 - 585) = 585/1115 = 0.525

(b). Heat expelled to the surrounding (room) = 1750J

coil wire, increase current, put iron inside the coil

the simple machine has three main features. they are:

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever