Question

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 1

The initial temperature of water is : 370.19 K

Given information : Mass of water = 587.00 Kg

Energy = 87 KJ

Specific heat of water = 1 J/Kg.K

Final temperature = 518.4 K

Initial temperature is unknown. Let take it as 'x'

Initial temperature can be calculated using the specific heat formula which is :

$Q = m\times c\times \triangle T$

$\triangle T = Final T - Initial T$

$Q = m\times c\times (Final T - Initial T)$

Where, Q = energy

m = mass

c = specific heat

$\triangle T$ = Change in temperature

We can plug in the values of Q, m , c and final T in the formula and can find the value of initial T.

Before plugging the values in the formula we need to look for the units. Since the unit of specific heat (c) is in J/Kg.K , we need to change the unit of energy from KJ to J

Energy = 87 KJ

$87 KJ \times \frac{1000J}{1KJ}$

Energy(Q) = 87000 J

$Q = m\times c\times (Final T - Initial T)$

$87000 J = 587.00 Kg \times \frac{1J}{Kg.K}\times (518.4 - x)$

$87000 J = \frac{587 Kg.J}{Kg.K}\times (518.4 -x)$

Same unit will be cancel out , in above step 'Kg' unit gets cancel out.

$87000 J = \frac{587J}{K}\times (518.4 -x)$

On rearranging the above equation we will get:

$\frac{87000J.K}{587 J} = (518.4 K -x)$

148.2112 K = 518.4 K - x

x = 518.4K - 148.2112K

x = 370.19 K

And 'x' we taken as initial temperature , so our initial temperature is 370.19 K