# Question 54a28

Nov 6, 2017

$\text{12,000,000 J}$

#### Explanation:

The equation that you need to use here looks like this

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{q = m \cdot c \cdot \Delta T}}}$

Here

• $q$ is the heat lost or gained by the water
• $m$ is the mass of water
• $c$ is the specific heat of water, equal to ${\text{4.184 J g"^(-1)""^@"C}}^{- 1}$
• $\Delta T$ is the change in temperature

Now, you know that your sample of water has a mass of $\text{95 kg}$. Right from the start, the fact that the specific heat of water is measured in joules per gram Celsius tells you that you need to convert the mass of water from kilograms to grams.

95 color(red)(cancel(color(black)("kg"))) * (10^3color(white)(.)"g")/(1color(red)(cancel(color(black)("kg")))) = 95 * 10^3color(white)(.)"g"#

You also know that the temperature of the water must increase by ${30}^{\circ} \text{C}$, so you can say that the change in temperature is equal to ${30}^{\circ} \text{C}$.

This basically means that you need to assume that the water starts at a given temperature as liquid water and ends at a temperature that is ${30}^{\circ} \text{C}$ warmer as liquid water, i.e. that you're not dealing with a phase change here.

So all you have to do now is to simply plug in your values into the equation and find the amount of heat needed for your sample.

$q = 95 \cdot {10}^{3} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * "4.184 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * 30color(red)(cancel(color(black)(""^@"C}}}}$

$q = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{12,000,000 J}}}}$

I'll leave the answer rounded to two sig figs, but keep in mind that you have one significant figure for the change in temperature.