# How many joules are required to heat 250 grams of liquid water from 0°C to 100°C?

May 28, 2016

$Q = 104525 J$

#### Explanation:

The key equation is

$Q = m c \Delta T$,

where $Q$ is the heat energy required to heat a mass, $m$,
though a change in temperature of $\Delta T$ and $c$ is the heat capacity of the material (found in books!). Typical units are $Q$ in Joules or $J$, $m$ in $k g$ and $\Delta T$ is in either Kelvin, $K$, or degrees Celsius ""^oC. The units of $c$ depend on the other units used but $J / k g$ ""^o C is typical.

we look up $c$ for water (textbook, the internet, etc)
$c = 4181$ $J / k g$ ""^o C
$m = 250$ $g = 0.250$ $k g$
$\Delta T = {T}_{\text{final" -T_"initial}} = 100$ ""^oC-0 ""^oC=100 ""^oC

Q=(0.250 kg)(4181 J // kg ""^o C) (100 ""^o C)
Q=(0.250 cancel{kg})(4181 J // cancel{kg} cancel{""^o C}) (100 cancel{""^oC})

$Q = 104525 J$