Question

The entire $8400J$ energy is converted into heat to raise the temperature of $1kg$ of water. What will be the rise in temperature? The specific heat of water is $4200Jk{g}^{-1}°C^{-1}$.

Answer 1

Step 1: The Given Data

Heat supplied,

$Q=8400J$.

Mass of water,

$m=1kg$

Specific heat of water,

$c=4200Jk{g}^{-1}°C^{-1}$

Step 2: Finding rise in temperature $△t=?$

Heat supplied is given as,

$$\begin{gathered} Q=mc△t \\ \Rightarrow △t=\frac{Q}{mc} \\ =\frac{8400J}{1kg\times 4200Jk{g}^{-1}°C^{-1}} \\ =\frac{8400}{4200} \\ =2°C \end{gathered}$$

Hence, temperature will rise by $2°C$.