# 1.2 xx 10^5 J of thermal energy are added to a sample of water and its temperature changes from 229 K to 243 K. What is the mass of the water?

Mar 23, 2016

$\Delta H = m C \Delta T$

Rearranging:

$m = \frac{\Delta H}{C \Delta T} = \frac{1.2 \times {10}^{5}}{4181 \times \left(243 - 229\right)} = 2.1$ $k g$

#### Explanation:

We use the equation $\Delta H = m C \Delta T$ where $m$ is the mass $\left(k g\right)$, $T$ is the temperature $\left(K\right)$, $H$ is the thermal energy $\left(J\right)$ and $C$ is the specific heat $\left(J k {g}^{-} 1 {K}^{-} 1\right)$. $\Delta$ just means 'the change in'.

We know that for water, $C = 4181$ $J k {g}^{-} 1 {K}^{-} 1$. (Sometimes quoted as $4.2$ $J {g}^{-} 1 {K}^{-} 1$, but it's better to work in SI units.)