Question

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

Answer 1

Q = 2926 kJ

Step-by-step 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* 4180* (35-0)`

Q = 2926000 Joules

or

Q = 2926 kJ

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