Calculating Heat Transfer and Work in Cyclic Processes | Refrigerator Example

[ariana0923]

Homework Statement

Over several cycles, a refrigerator does 1.51 x 10^4 J of work on the refrigerant. The refrigerant in turn removes 7.55 x 10^4 J as heat from the air inside the refrigerator.

a. how much energy is transferred as heat to the outside air?
b. what is the net change in the internal energy of the refrigerant?
c. what is the amount of work done on the air inside the refrigerator?
d. what is the net change in the internal energy of the air inside the refrigerator?

Homework Equations

U=Internal Energy
Q=Heat
W=Work

ΔU = Q-W
ΔUnet = 0
Qnet=Wnet

The Attempt at a Solution

I'm trying to teach myself physics, so I think I'm getting really confused here. First of all, I get mixed up on when to put negative signs in front of numbers or if I even have to in cyclic processes.

I got the right answers (given by my online school) but it may have just been luck.

**I only need help with a and d and somewhat ca. -7.55 x 10^4 = (1.51 x 10^4 - x)
-9.06 x 10^4= -x
x=9.06 x 10^4 J

Did i do that right at all?

b. I know ΔUnet=0 J in cyclic processes

c. no work done b/c ΔV=0 (how would i know that this is so?)

d. the answer is ΔU= -7.55 x 10^4 but I'm not sure how to mathematically get it

 

[queenofbabes]

a) If you remove heat from the fridge, you have to dump it outside. It should just be 7.55*10^4 J

d) Internal energy is just a function of temperature. You remove heat, thus its considered a negative change.

 

[kuruman]

c) When there is no volume change, there is no displacement of anything due to the force (pressure times area) exerted by the gas. Therefore the gas does no work when its volume is constant.