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
