A sample of water with a mass of 587.00 kg is heated with 87 kJ of energy to a temperature of 518.4 K. The specific heat of water is 1 J-1 kg K-1. What is the initial temperature of the water?

Answer : Initial temperature of the water is 518.3 K

Explanation:

The specific heat of water (Cp) can be expressed in 2 ways

1) Cp = 1 J g⁻¹ K⁻¹ or

2) Cp = 1 kJ kg⁻¹ K⁻¹

The amount of heat absorbed by water is given by the following equation.

Q = m*Cp*ΔT

Here Q is the amount of energy absorbed in kilojoules

m is the mass of water in kilograms

Cp is specific heat of water

ΔT is the difference is temperature of water.

We have been given that

Q = 87 kJ

Cp = 1 kJ kg⁻¹ K⁻¹

m = 587.00 kg

Let us plug in these values in equation for Q

Q = mxCpxΔT

87 kJ = 587.00 kg x 1 kJ kg⁻¹ K⁻¹ x ΔT

87/587.00 = ΔT

ΔT = 0.15 K

ΔT is calculated as Tf - Ti where Tf is the final temperature and Ti is the initial temperature.

We know that Tf = 518.4 K

And we calculated ΔT as 0.15 K.

Let us plug these values in the equation

ΔT = Tf - Ti

0.15 K = 518.4 K - Ti

Ti = 518.25 K

Initial temperature of the water is 518.3 K