# What is the final temperature of 400 g of water at 20°C after it absorbs 226 kJ of heat?

May 26, 2016

I found: ${T}_{f} = {155}^{\circ} C$

#### Explanation:

I would use the relationship:
$Q = m c \Delta T$
where:
$Q$ is the heat;
$m$ is the mass;
$c$ is the specific heat;
$\Delta T = {T}_{f} - {T}_{i}$ is the difference in temperature.
From literature we have:
${c}_{\text{water}} = 4.18 \frac{J}{{g}^{\circ} C}$
So we have:
$226 \times {10}^{3} = 400 \cdot 4.18 \left({T}_{f} - 20\right)$
${T}_{f} = {155}^{\circ} C$