A solar heated house loses about 5.4 × 107 cal through its outer surfaces on a typical 24-h winter day.

Physics

1 answer:

mojhsa [17]2 years ago

4 0

Answer:

Explanation:

Q=mcΔθ

Q=quantity of heat , m= mass of the storage rock

Δθ= temperature change.

m= Q/(cΔθ)

Q=5.4 $10^{7}$

Δθ=62°C-20°C

=42°C

c=0.21cal/g.°C

$m=\frac{5.4*10^{7} }{0.21*42} \\\\m=6122448.98g\\$

m≈6100000g

m≈6100kg

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