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! :)

 

[asteriatic]

I would approach this problem by first understanding the concept of entropy and its relationship to heat transfer. Entropy is a measure of the disorder or randomness in a system, and it increases with heat transfer from a warm to a cool environment. In this case, the house is losing heat to the outside, so the entropy of the system is increasing.

To calculate the change in entropy, we can use the equation dS = dQ/T, where dQ is the heat transfer and T is the temperature in Kelvin. However, we need to be careful in defining the boundaries of our system. The question mentions the temperature inside the house, Tin, and the temperature outside, Tout. This suggests that the system includes both the house and the outside environment.

Using this approach, we can calculate the change in entropy as follows:

dS = -5.49 kJ / 283.7 K + 5.49 kJ / 283.7 K = 0

This means that the change in entropy caused by the heat leaking from the house is zero. This may seem counterintuitive, but it makes sense because the heat transfer is occurring between two systems at different temperatures, and the total entropy change is the sum of the entropy changes in each system. In this case, the decrease in entropy inside the house is equal to the increase in entropy outside, resulting in a net change of zero.

In summary, when calculating the change in entropy caused by a heat leak from a house, it is important to consider the entire system, including both the house and the outside environment.