Question

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

What mass of storage rock is needed to provide this amount of heat if it is brought up to initial temperature of 62°C by the solar collectors and the house is maintained at 20°C? (Specific heat of rock is 0.21 cal/g⋅°C.)

a. 163 kg
b. 1 230 kg
c. 6 100 kg
d. 12 700 kg

Answer 1

C

Step-by-step 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=(5.4*10^(7) )/(0.21*42) \\\\m=6122448.98g\\`

m≈6100000g

m≈6100kg