Compute work in Carnot Refrigerator

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1. Feb 9, 2016

duran9987

1. The problem statement, all variables and given/known data
The temperature inside the room $T_I= 25 C$ and the temperature outside the house is $T_O = 32C$. The temperature difference causes energy to flow into the room (by conduction through the walls and window glass) at the rate 3,000 J/s. To return this energy to the outside by running an ideal refrigerator, how much electrical energy must be supplied to the refrigerator (to perform the external work)?

2. Relevant equations
$Q_h=W_{in}+Q_c$, Energy Conservation
$Efficiency = 1 - T_c/T_h=(Q_h-Q_c)/Q_h$

3. The attempt at a solution
My thinking is that we have to put in work so that the refrigerator takes energy from the cold reservoir and dumps 3000 Joules onto the hot reservoir. This leads me to set $Q_h=3000$ and use the efficiency to find $Q_C$ and eventually solve for $W_{in}$.

$Efficiency = 1 - (298.15K/305.15K) = 0.02264$

$0.02264 = (3000-Q_c)/3000 ⇒ Q_c = 2931.19J$

$W_{in} = 3000J-2931.19J = 68.82J$

Is my approach correct? seems like a small amount of energy needed to dump 3000J back to the outside. Any feedback (or more efficient methods) would be greatly appreciated.

2. Feb 9, 2016

TSny

Your approach looks pretty good. However, are you sure that 3000 J should be used for $Q_h$ rather than for $Q_C$?

3. Feb 9, 2016

duran9987

I chose $Q_h$ because the problem wants me to return 3000 joules of energy to the outside. If I set $Q_c$ to 3000J I would be returning more energy (Including the energy from the work, which wouldn't be wrong i guess since I technically am returning the incoming 3000 joules plus some extra). I guess either way I can't be wrong if the math checks out.

4. Feb 9, 2016

TSny

Perhaps it's open to interpretation. But, it seems to me you want to keep the room from continually warming up. So, for every 3000 J that enters the room from outside, you need to remove 3000 J from the room. You will then be returning more than 3000 J to the outside. (You will also have to position the refrigerator such that the back of the refrigerator is outside the house while the front of the refrigerator is inside the house with the door of the refrigerator kept open!)