Question

Water (2850 g ) is heated until it just begins to boil. If the water absorbs 5.53×105 j of heat in the process, what was the initial temperature of the water?

Answer 1

Answer is 54 °C.

Explanation;

We can simply use heat equation

Q = mcΔT

Where Q is the amount of energy transferred (J), m is the mass of the substance (kg), c is the specific heat (J g⁻¹ °C⁻¹) and ΔT is the temperature difference (°C).

Let's assume that the initial temperature is T.

Q = 5.53 × 10⁵ J

m = 2850 g

c = 4.186 J/g °C

ΔT = (100 - T) °C Since the water is boiling, the final temperature is 100 °C.

By applying the equation,

5.53 × 10⁵ J = 2850 g x 4.186 J/g °C x (100 - T) °C

(100 - T) °C = 5.53 × 10⁵ J / (2850 g x 4.186 J/g °C )

(100 - T) °C = 46.35 °C

T = 100 - 46.35 C = 53.65 °C

≈ 54 °C