Question

How much heat energy is required to warm up 2.35 kg of water, from 20oC to 83oC ?

Answer 1

the Given data

The mass of the water is m = 2.35 kg

The initial temperature of water is T1 = 20oC

The final temperature of the water is T2 = 83oC

The expression for the heat energy required to warm up is given as:

`q=mC(T_2-T_1)`

The specific heat capacity of water is given as:

`C=^{}4186J/kg^o\text{C}`

Substitute the value in the above equation.

`\begin{gathered} q=2.35\text{ kg}*4186J/kg^o\text{C }*(83^(\circ)C-20^(\circ)C) \\ q=619737.3\text{ J} \\ q=619737.3\text{ J}*\frac{1\text{ kJ}}{1000\text{ J}} \\ q=619.737\text{ kJ} \end{gathered}`

Thus, the amount of heat energy required to warm-up is 619.737 kJ.