When power was turned off to a 30.0 gal. water heater, the temperature of the water dropped from 75.0°C to 22.5°C. How much heat was lost to the surroundings?

Question

Quizplus · Accepted Answer

We can use the equation Q = mcΔT, where Q is the heat lost to the surroundings, m is the mass of water (which we can calculate from the volume and density of water), c is the specific heat capacity of water, and ΔT is the change in temperature.

First, let's calculate the mass of water:
density of water = 1.00 g/mL
30.0 gal = 113.56 L

mass = volume x density = 113.56 L x 1000 g/L = 113560 g

Now we can plug in the values for Q, m, c, and ΔT:

Q = (113560 g) x (4.18 J/g°C) x (-52.5°C)
Q = -2.49 x 10^14 J
(Note that the temperature difference is negative because the water is cooling down.)

Converting to kilojoules gives:
Q = -2.49 x 10^14 J ÷ 1000 J/kJ = -2.49 x 10^11 kJ

So the best choice is C.

When Power Was Turned Off to a 30