Finding entropy change when house leaks heat

[rubenhero]

Homework Statement

b) The temperature inside a house is Tin = 19.2° C when the temperature outside is Tout = 10.7° C. If Q = 5.49 kJ of heat leak from the house to the outside, find the change in entropy caused by this process.

Homework Equations

W = Q_H - Q_C, dS = dQ/T

The Attempt at a Solution

dS = 5.49kJ/283.7K = .0193514276kJ/K

The answer I calculated was wrong, should I include the entropy inside the house?
Any help is appreciated!

 

[I like Serena]

Hi rubenhero!

Homework Statement

b) The temperature inside a house is Tin = 19.2° C when the temperature outside is Tout = 10.7° C. If Q = 5.49 kJ of heat leak from the house to the outside, find the change in entropy caused by this process.

Homework Equations

W = Q_H - Q_C, dS = dQ/T

The Attempt at a Solution

dS = 5.49kJ/283.7K = .0193514276kJ/K

The answer I calculated was wrong, should I include the entropy inside the house?
Any help is appreciated!

Inside you lose ΔS=-Q/T_H.
Outside you gain ΔS=Q/T_L.

The question in your problem statement is not entirely clear, but I suspect the total change in entropy is intended, which is the sum of both entropy changes.

 

[rubenhero]

Thanks you for responding. You were right, I just did the sum of the entropy loss and gain and got an answer of .5629265733 J/K. Webassign marked my answer as right. Thanks again for your help!

 

[I like Serena]

You're welcome! :)