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 :

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

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.

$87000J=648.00kg\times 1J/kg.K\times (T_{final}-298K)$

$T_{final}=432.26K$

Therefore, the final temperature of water is, 432.26 K