Question

If 40.0 kj of energy are absorbed by .500 kg of water 10 degrees c, what is the final temperature of the water?

Answer 1

To solve the problem, we must use the following equation:

$Q=mC_s (T_f -T_i)$

where

Q is the amount of heat energy absorbed by the water

m is the mass of the water

Ti and Tf are the initial and final temperature

Cs is the specific heat capacity of the water

The data we have in this problem are:

Q=40.0 kJ

$C_s =4.186 kJ/kg^{\circ}C$

$T_i=10^{\circ}C$

m=0.500 kg

Substituting the data into the equation and re-arranging it, we find

$T_f = T_i + \frac{Q}{mC_s}=10^{\circ}+\frac{40.0 kJ}{(0.500 kg)(4.186 kJ/kg^{\circ}}=29.1^{\circ}C$

So the final temperature of the water will be 29.1 degrees.