A solar heated house loses about 5.5 × 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 58°C by the solar collectors and the house is maintained at 23°C? (Specific heat of rock is 0.21 cal/g⋅°C. )

Question

Quizplus · Accepted Answer

We can use the formula:

Q = mcΔT

where Q is the amount of heat needed, m is the mass of the storage rock, c is the specific heat, and ΔT is the temperature change.

We can first convert the heat loss to Joules:

5.5 × 10^17 cal = 5.5 × 10^17 × 4.184 J/cal = 2.3 × 10^18 J

Next, we can calculate the temperature difference:

ΔT = (58°C - 23°C) = 35°C

Now we can solve for the mass:

m = Q / (cΔT) = (2.3 × 10^18 J) / (0.21 cal/g°C × 4.184 J/cal × 35°C) ≈ 7.5 × 10^3 kg

Therefore, the best answer is (C) 7.5E+3 kg.

A Solar Heated House Loses About 5