# Homework Help: Power consumption costs of a fridge with bad lights

1. Apr 5, 2016

### reed2100

1. The problem statement, all variables and given/known data
Note - I've asked a very similar problem to this before but I'm redoing it for practice, and this one asks a new question which just ruins everything. I also don't know if I was right the first time yet because I haven't gotten the paper back.

Consider a refrigerator whose 40-W light bulb remains on continuously as a result of a malfunctioning of the switch inside the refrigerator. The refrigerator has a coefficient of performance (COP) of 1.3 and the cost of electricity is estimated to be 10 cents per kWh (kilowatt-hour). Assume that the refrigerator is normally opened 20 times a day for an average of 30 s.

1. Determine the total additional power consumed by the refrigerator (15 pts).

2. Calculate the additional electric power consumption of the refrigerator (10 pts).

3. Estimate the additional electric power costs per year due to such malfunctioning per year (10 pts).
Note - I have no idea how 1 and 2 are any different, that's the crux of this submission. They both look exactly the same to me. Confused. Don't bother with the 3rd question, that part is easy enough to work out once I make better sense of the first two, and it will require those number(s) anyway.

2. Relevant equations
Power = work / time = Joule / second = Watts
COP = β = QL per sec / Wper sec

3. The attempt at a solution

COP = 1.3 , but I'm not given any baseline values for the heat the fridge takes out of the food space, what work input it requires to do so, or how much heat it rejects. The bulb is a 40 watt bulb, but I'm unclear what to assume about that here. Does that only mean it gives off 40 watts of heat the fridge then has to suck out, or does it also mean the bulb consumes 40 watts to function? So there are two potential things to be aware of with the bulb - when it is on the total power consumption increases by 40 watts to power it (unclear?), and the work input of the fridge will have to increase by a certain amount in order to boost the rate at which it sucks heat out of the food space by 40 watts.

So expressing this in numbers, looking at the 2nd issue -
β = 1.3 = QL per sec / Wper sec
1.3 = heat we're looking at per work input required for it
thus we can say = extra 40 watts of heat to suck out / work required to do so
40 / 1.3 = work required = 30.76 watts of extra work input to the fridge for it to suck out the heat produced by the light bulb when it's on

Now, if the only thing I have to worry about from the lightbulb in terms of power is the extra power consumed by the fridge to counteract its heat, then I would say the additional power consumption of the fridge during the time the bulb is on is 30.76 watts.

But if that's not the case, and the bulb essentially consumes 40 watts to function, then I would say the total additional power consumption when the bulb is on is actually "extra power to offset the heat + power to power the bulb" , which

= (30.76 + 40) watts
= 70.76 watts

That's all I have to work with right now, I'm confused on the question wording.
I don't really understand how questions A and B are any different, they sound vague already. They don't specify any length of time so the answers must be in watts. But " determine total additional power consumed by fridge" and "calculate additional electric power consumption of the fridge" literally sound identical to me. I don't know if the inclusion of the word "electric" means anything important here or what, it's all power, all the same units. The heat from the bulb necessitates an extra work input to the fridge, and the bulb (unclear?) requires 40 watts to function. Am I just missing an understanding of terminology or something?

Any viewpoints / insight and help is greatly appreciated, thanks.

2. Apr 5, 2016

### Staff: Mentor

Both.

The fridge is opened sometimes, where the light bulb is designed to be on. That changes the answers a bit.
I don't see the difference between 1 and 2, unless you subtract emitted heat from the total power (which would be odd).

3. Apr 5, 2016

### reed2100

I see, thank you. I went with that and he just posted the solutions, I got the correct answers. A) 30.76 watts to keep the fridge temperature constant, B) 70.76 watts to power the bulb and compensate for the heat, and then after calculations C) \$61.60 a year extra to pay for it. I guess he could have just worded it more specifically.