# How much heat is required to warm 1.60 L of water from 26.0°C to 100.0°C?

May 23, 2016

495385.6 $J$ or 495.4 $k J$

#### Explanation:

To solve this problem, we need to know/look up this formula :

$q = m \times \Delta T \times c$

Where:

$q$ - amount of heat
$m$ - mass
$c$ - specific heat capacity
$\Delta T$ - Temperature difference

Given:
V = 1.60L
$\Delta T$ = (100-26 = 74)
Knowing that the liquid is water, we can use density to get the mass.
$D$ of water = 1.0g/mL

1.60L x 1.0g/mL x 1000mL/1L = 1600g (We used 1000mL/1L conversion to get rid of any Volume Units)

Specific heat capacity is either given or you have to look this up also. Anyway, in this case we are dealing with water in a liquid phase. Therefore, $c = \text{4.184 J/g"cdot""^@"C}$

Now all you have to do is plug everything in the formula above.

$q$ = 1600g x (100-26C) x ($\text{4.184 J/g"cdot""^@"C}$) = 495385.6 $J$ or 495.4 $k J$