# Refrigerator Light (power and Carnot engines)

1. Mar 14, 2007

### bugonwndshld

The inside of an ideal refrigerator is at a temperature , while the heating coils on the back of the refrigerator are at a temperature . Owing to a malfunctioning switch, the light bulb within the refrigerator remains on when the the door is closed. The power of the light bulb is ; assume that all of the energy generated by the light bulb goes into heating the inside of the refrigerator.

For all parts of this problem, you must assume that the refrigerator operates as an ideal Carnot engine in reverse between the respective temperatures.
If the temperatures inside and outside of the refrigerator do not change, how much extra power does the refrigerator consume as a result of the malfunction of the switch? (Express the extra power in terms of P,T_h,and T_c)

i know that Q/W_in = Tc/T_h-T_c
and W_out/Q_H = 1-T_c/T_h

i'm just not sure how to equat this to power... would the W_in be the power of the light bulb?

i get something like 1-(P(T_c/T_h-T_c) but that can't be right...

if someone could help, that would be great!

2. Mar 14, 2007

### G01

If the power of the light bulb is x Joules/s, then consider the refrigerator over and interval of 1 second. The refrigerator must move x Joules from the cooler in that second to keep up with the light bulb. using the value for x, and the efficiency of the engine, which you can find from the hot and cold temperatures for the Carnot engine, you can find how much extra work the refrigerator did in order to compensate for the bulb. Divide this work by the time span(1 second), and you have the extra power!

3. Mar 15, 2007

### bugonwndshld

ok... so the efficiency of a carnot engine is 1-(T_c/T_h), correct? so, the power, P, is W_out=P_out(delta T)/P_h (delta T)
so, wouldn't the equation be (P/(1-(T_c/T_h)))-P ??
i'm sorry, i still don't see what i'm doing wrong...

4. Mar 15, 2007

### G01

This is correct.

Now, call the energy needed to be dumped into the hot reservoir, Q_hot. This is equal to the energy given off by the light bulb during the span of 1 second. Now remember for any engine or refrigerator:

$$eff=\frac{W}{Q_{hot}}$$

Now using this information and what you already know about the efficiency, can you find the total energy needed ? After that, then divide this energy by the time interval(1 sec) to find the extra power used.

See how far you can get now. Good Luck!

Last edited: Mar 15, 2007
5. Mar 15, 2007

### bugonwndshld

ok, lets see... this is really frustrating me!

so if

so if (1-(T_c/T_h)=W/Q_hot, then Q_hot= W/(1-(T_c/T_h)) (that is the energy given off by the light bulb) so it's W+(W/(1-(T_c/T_h))? is that right? how do you find W?

6. Mar 15, 2007

### G01

I do not know where this line came from:

W+(W/(1-(T_c/T_h))

EDIT: Did you create this equation because your finding the TOTAL work done by the engine, as in the normal work done plus the extra work. If this is the case, the question you posted asks only for the extra power, in which case you only need to find the extra work done.

This is correct:

Look at the second equation in the quote above.

Can't you solve that for the work done by the engine?

Last edited: Mar 15, 2007
7. Mar 15, 2007

### bugonwndshld

W= Q_hot(1-(T_c/T_h)

that is the work done by the engine? is the Q_hot the wattage of the light bulb? and if you divide this by the time interval, 1 second, it wouldn't change the equation, right?

oh my gosh, i'm sorry! i'm not very good at physics!

8. Mar 15, 2007

### G01

No problem, I think that is it, except Q_hot is the energy of the light bulb, divide both sides by the time interval to get the power, or wattage. All that does is change W to the Extra Power, call it P_extra, and Q_hot to the power of the light bulb, call it P_bulb.

Your not bad at physics! Everyone struggles at times, even the best of the best!
Good luck in the future!

9. Mar 15, 2007

### bugonwndshld

ooh, i got it! the answer was -P(1-(T_h/T_c)), we needed to switch the c and h because we were dividing... yay!

thanks so much for your help!

10. Mar 15, 2007