Question

Water (3030 g ) is heated until it just begins to boil. If the water absorbs 5.49×105 J of heat in the process, what was the initial temperature of the water?

Answer 1

The initial temperature of the water was 56.7 °C.

Step 1. Calculate the temperature change.

The formula for the heat (q) absorbed is

q = mCΔT

We can solve this equation to get

ΔT = q/(mC)

q = 5.49 × 10⁵ J; m = 3030 g; C = 4.184 J·°C⁻¹g⁻¹

∴ ΔT = 5.49 × 10⁵ J/(3030 g × 4.184 J·°C⁻¹g⁻¹) = 43.30 °C

Step 2. Calculate the initial temperature

ΔT = T_f – T_i = 100°C – T_i = 43.30 °C

T_i = 100 °C – 43.30 °C = 56.7 °C