# Can you go over q = m * c * DeltaT ?

Sep 8, 2014

The specific heat capacity, or simply specific heat $\left(C\right)$ of a substance is the amount of heat energy required to raise the temperature of one gram of the substance by one degree Celsius. Heat energy is usually measured in Joules $\left(\text{J}\right)$ or calories $\left(\text{cal}\right)$.

The variables in the equation $q = m C \Delta T$ mean the following:

$\text{ let:}$
$q = \text{heat energy gained or lost by a substance}$
$m = \text{mass (grams)}$
$C = \text{specific heat}$
$\Delta T = \text{change in temperature}$

Note that $\Delta T$ is always calculated as $\text{final temperature "-" initial temperature}$, not the other way around.

Therefore, you can look at that equation like this if it helps:

"heat energy gained or lost by a substance"=("mass")("specific heat")(DeltaT)

Example
How much heat energy is required to raise the temperature of $\text{55.0g}$ of water from $\text{25.0"^@"C}$ to $\text{28.6"^@"C}$? The specific heat of water is $\text{4.18""J}$/$\text{g"^@"C}$. This is a very well known specific heat value and will frequently show up in specific heat questions.

Unknown:
$q$ in Joules $\left(\text{J}\right)$

Known/Given:
specific heat of water $\left(C\right)$ = $\text{4.18""J}$/$\text{g"^@"C}$
mass of water = $\text{55.0g}$
$\Delta T$ = $\text{28.6"^@"C"-"25.0"^@"C}$ = $\text{3.6"^@"C}$

Equation:
$q = m c \Delta T$

Solution:
$q = \text{55.0g" xx "4.18""J}$/$\text{g"^@"C"xx"3.6"^@"C}$
$q$ = $\text{827.64""J}$, which rounds to $8.28 \times {10}^{2} \text{J}$ due to significant figures.