What is the Rate of Entropy Production from Heat Leakage in a House?

[Bill Foster]

Homework Statement

Heat leaks out of a house at the rate of $$2.5\times{10^4} \frac{kcal}{h}$$. The temperature inside the house is 21°C and the temperature outside the house is -5°C. At what rate does this process produce entropy.

Homework Equations

$$S(A)-S(A_0)=-\frac{\Delta{Q}}{T_1}+\frac{\Delta{Q}}{T_2}=\Delta{Q}\times{(\frac{1}{T_2}-\frac{1}{T_1})}$$

The Attempt at a Solution

Heat flows from the high temp res to the low temp res at a rate of $$2.5\times{10^4} \frac{kcal}{h} = 6944.4 \frac{cal}{s} = \Delta{Q}$$.

$$T_1=21°C=294.15K$$
$$T_1=-5°C=268.15K$$

Plugging and chugging I get: $$2.29 \frac{cal}{K\times{s}}$$, which is wrong.

The answer in the back of the book is $$8.2\times{10^3}\frac{cal}{K\times{s}}$$.

 

[hage567]

If you leave your answer in cal/(K x h), you get the book's answer. Maybe they made a mistake?

 

[Bill Foster]

Now I feel stupid for overlooking a such a trivial detail as unit conversions.