# If it takes 21,966 J of heat energy to warm 750 g of water, what was the temperature change?

Apr 18, 2017

The temperature change was ${7.0}^{\circ} \text{C}$.

#### Explanation:

Use the equation:

$q = m c \Delta t$

where $q$ is the heat gained or lost in Joules, $c$ is specific heat capacity, $m$ is mass, and $\Delta t$ is the temperature change, which is $\left({T}_{\text{final"-T_"initial}}\right)$.

The specific heat of water, $\left({c}_{\text{H2O}}\right)$ is 4.184 "J"/("g"*^@"C")
https://water.usgs.gov/edu/heat-capacity.html

Organize the Information

Given/Known
c_"H2O"=4.184 "J"/("g"*^@"C")
$m = \text{750 g}$
$q = \text{21966 J}$

Unknown: $\Delta t$

Solution
Rearrange the equation to isolate $\Delta t$. Substitute the given/known values into the equation and solve.

$\Delta t = \frac{q}{m \times c}$

Deltat=(21966color(red)cancel(color(black)("J")))/(750color(red)cancel(color(black)("g"))xx(4.184 color(red)cancel(color(black)("J"))/(color(red)cancel(color(black)("g"))*^@"C")))="7.0"^@"C"