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

Answer:

C

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