A solar heated house loses about 5.4 × 107 cal through its outer surfaces on a typical 24-h winter day.
1 answer:
Answer:
C
Explanation:
Q=mcΔθ
Q= quantity of heat, m= mass of the storage rock
Δθ= change in temperature.
Solving for m: m= Q/(cΔθ)
Where Q=5.4
Δθ=62°C-20°C
=42°C
c=0.21cal/g.°C
Calculating gives us m≈6100000g
m≈6100kg
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