# Question 6ad15

Mar 3, 2017

$\text{21,000 J}$

#### Explanation:

You can find the amount of heat needed to increase the temperature of your sample by

$\Delta T = {50}^{\circ} \text{C}$

by using the equation

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

Here

• $q$ is the heat lost or gained by the substance
• $m$ is the mass of the sample
• $c$ is the specific heat of the substance
• $\Delta T$ is the change in temperature, defined as the difference between the final temperature and the initial temperature of the sample

Now, the trick here is to realize that the mass of the sample is given to you in grams, but that the specific heat of water is expressed in joules per kilogram Celsius.

This means that you will have to convert the mass of the sample from grams to kilograms by using the fact that

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{1 kg" = 10^3color(white)(.)"g}}}}$

In your case, you will have

100 color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = "0.1 kg"#

Now you're ready to plug your values into the equation and solve for $q$

$q = 0.1 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{kg"))) * "4,184 J" color(red)(cancel(color(black)("kg"^(-1))))color(red)(cancel(color(black)(""^@"C"^(-1)))) * 50 color(red)(cancel(color(black)(""^@"C}}}}$

$q = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{21,000 J}}}}$

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