# If 335 g of water at 65.5°C loses 9750 J heat, what is the final temperature of the water?

Mar 26, 2017

The final temperature is =58.6ºC

#### Explanation:

Let the final temperature be =TºC

Then,

DeltaT=(65.5-T)ºC

Mass of water, $m = 0.335 k g$

Heat lost, $H = 9.750 k J$

specific heat of water, s=4.186kJkg^-1ºC^-1

Therefore,

$H = m s \Delta T$

$9.750 = 0.335 \cdot 4.186 \cdot \left(65.5 - T\right)$

$65.5 - T = \frac{9.750}{0.335 \cdot 4.186} = 6.95$

T=65.5-6.95=58.6ºC