# What is the temperature change in 224 g of water upon the absorption of 55 kJ of heat, the specific heat of water is 4.18 J/g °C?

Dec 18, 2016

$\Delta T = {58.7}^{o} C$

#### Explanation:

To obtain the change in temperature, let's use the equation below: In this question, heat has units of kJ, which does not match up with the units given in the image above. So what we can do first is convert kJ to J by multiplying that value of $q$ by 1000 using the conversion factor below: Here are the variables that we know:

$Q$ = 55cancel"kJ"xx(1000J)/(1cancel"kJ") = 55,000 J
$m$ = 224 g
$C$ = $4.18 \frac{J}{g {\times}^{o} C}$

All we have to do is rearrange the equation to solve for $\Delta T$. This can be accomplished by dividing both sides by $C$ and $m$ to get $\Delta T$ by itself like this:

$\Delta T = \frac{Q}{m \times C}$

Now, we just plug in the known values:

$\Delta T = \frac{55 , 000 \cancel{J}}{224 \cancel{g} \times 4.18 \frac{\cancel{J}}{\cancel{\text{g}}} {\times}^{o} C}$

$\Delta T = {58.7}^{o} C$