Question

a sample of water with a mass of 648.00 kg at 298 K is heated with 87 kh of energy. the specific heat of water is 1 J-1 kg K-1. what is the final temperature of the water

Answer 1

q = mCΔT

The correct specific heat capacity of water is 4.187 kJ/(kg.K).

ΔT = q/mC = 87 kJ/[648.00 kg x 4.187 kJ/(kg.K)] = 87 kJ/(2713 kJ/K) = 0.032 K

Tf = Ti + ΔT = 298 K + 0.032 K = 298.032 K

Answer 2

Answer : The final temperature of water is, 432.26 K

Solution :

Formula used :

`Q=m* c* \Delta T=m* c* (T_(final)-T_(initial))`

where,

Q = heat supply = 87 kJ = 87000 J

m = mass of water = 648.00 kg

c = specific heat of water =
`1J/kg.K`

`\Delta T=\text{Change in temperature}`

`T_(final)` = final temperature = ?

`T_(initial)` = initial temperature = 298 K

Now put all the given values in the above formula, we get the final temperature of water.

`87000J=648.00kg* 1J/kg.K* (T_(final)-298K)`

`T_(final)=432.26K`

Therefore, the final temperature of water is, 432.26 K