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Scilla [17]
4 years ago
13

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]4 years ago
4 0

Answer:

C

Explanation:

Q=mcΔθ

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

Δθ= temperature change.

m= Q/(cΔθ)

Q=5.410^{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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