Question

How much heat energy is required to raise the temperature of 20 kilograms of water from 0°c to 35°c?

Answer 1

Energy required = Specific Heat × mass × change in temp. 
Specific Heat (Cp) water = 4.186kJ/kgC 
Mass = 20kg 
Change in temperature = 35C - 0C = 35C 

Q = 4.186kJ/kgC × 20kg × 35C = 293.02kJ     hope this helps! (:

Answer 2

Answer:

Q = 2926 kJ

Explanation:

Given that,

Initial temperature of water, $T_1=0^{\circ}$

Final temperature of water, $T_2=35^{\circ}$

Mass, m = 20 kg

To find,

Heat energy required to raise the temperature.

Solution,

Let Q is the heat energy required to raise the temperature. It is based on the concept of specific heat. The expression for the heat raised is given by :

$Q=mc\Delta T$

c is the specific heat of water, $c=4180\ kJ/kgC$

$Q=20\times 4180\times (35-0)$

Q = 2926000 Joules

or

Q = 2926 kJ

So, the heat energy required to raise the temperature of water is 2926 kJ.