How much heat is added if .617g of water is increased in temperature by .241 degrees C?

Jan 2, 2018

0.622 J

Explanation:

In order to solve this problem, you need the formula:
$q = m {C}_{s} \Delta T$

Assuming water is in its liquid state, the specific heat capacity or ${C}_{s}$ would be 4.18 $\frac{J}{g \cdot \mathrm{de} g r e e s C e l s i u s}$

$\Delta T$ in this case would be 0.241 degrees Celsius since this represents the change in temperature.

$m$ would be 0.617 grams.

Plug in all the values and you would get 0.622 joules!
Hope this helps!