Question

A home solar energy unit stores energy for later use. On a sunny day, the temperature of the water in the tank increased from 23.4naughtC to 39naughtC. If 2.55âââ104 kJ were absorbed by the water, what volume of water was in the tank? The density and specific heat of water are 0.998 g/mL and 4.184 J/gnaughtC. Round your answer to the nearest whole number.

Answer 1

391.46 L

Step-by-step explanation:

Given:

Initial temperature of the water = 23.4°C

Final temperature of the water = 39°C

Temperature change for the water, ΔT = (39 - 23.4) = 15.6°C

Heat absorbed by the water = 2.55 × 10⁴ kJ = 2.55 × 10⁷ J

Density of the water = 0.998 g/mL

Specific heat of the water, C = 4.184 J/g°C

Now,

The heat absorbed by the water, q = mCΔT

where, m is the mass of the water

Thus,

we have

2.55 × 10⁷ J = m × 4.184 × 15.6

or

m = 390682.45 grams

also,

Density = mass / volume

thus,

0.998 = 390682.45 / volume

or

Volume = 391465.38 mL

or

Volume = 391.46 L