# What is the number of joules of heat energy released when 20 grams of water is cooled from 293 K to 283 K?

Jun 21, 2017

836 J

#### Explanation:

Use the formula q = mCΔT

q = heat absorbed or released, in joules (J)
m = mass
C = specific heat capacity
ΔT = change in temperature

Plug known values into the formula.

The specific heat capacity of water is $4.18 \frac{J}{g} \cdot K$.

$q = 20 \left(4.18\right) \left(293 - 283\right)$
$q = 20 \left(4.18\right) \left(10\right)$
$q = 836$

836 joules of heat energy are released.