What is the final temperature after 840 J is absorbed by 10.0 g of water at 25°C?

May 15, 2016

The final temperature is equal to ${45}^{\circ}$ $C$

Explanation:

We have the amount of energy gain
$Q = m \cdot c \cdot \Delta T = m c \Delta t$

where $c$ = $4.184$ $J$/$g . C$ is the specific heat of water, $m$ is the mass of water

$\implies$$840 = 10$ x $4.184 \cdot \left(t - 25\right)$
$t = \frac{840}{10 \text{ x } 4.184} + 25 = 45$ i.e. ${45}^{\circ}$ $C$