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 $\degree$C by the solar collectors and the house is maintained at 20 $\degree$C? (Specific heat of rock is 0.21 cal/g*$\degree$C. ) A)163 kg B)1 230 kg C)6 100 kg D)12 700 kg

Question

Quizplus · Accepted Answer

We can use the formula Q = mcΔT to solve for the mass of storage rock needed.

Q = amount of heat lost = 5.4 × 10^17 cal 
m = mass of storage rock (unknown) 
c = specific heat of rock = 0.21 cal/g·°C 
ΔT = temperature change = (20 - 62)°C = -42°C

Solving for m:

5.4 × 10^17 cal = m × 0.21 cal/g·°C × (-42°C) 
m = 6.1 × 10^9 g = 6 100 kg

Therefore, the best choice is C.

A Solar Heated House Loses About 5 °\degree° C by the Solar Collectors and the House Is Maintained

A Solar Heated House Loses About 5 $\degree$ C by the Solar Collectors and the House Is Maintained