# What would be the final temperature if 3.31 * 10^3 joules of heat were added to 18.5 grams of water at 22.0°C?

Jan 30, 2016

I found: ${T}_{f} = {64.6}^{\circ} C$

#### Explanation:

We can try using the general relationship between heat $Q$ (lost or gained) and change in temperature $\Delta T$ of a mass $m$ of a substance of specific heat $c$ (in our case water: ${c}_{\text{water}} = 4.2 \frac{J}{{g}^{\circ} C}$):

$Q = m c \Delta T$

with our data:
$3.31 \times {10}^{3} = 18.5 \cdot 4.2 \cdot \left({T}_{f} - 22\right)$
rearranging we get:
${T}_{f} = {64.6}^{\circ} C$