# Question 5012c

Mar 11, 2017

$\text{3400 J}$

#### Explanation:

The first thing you need to do here is to look up the specific heat of water

${c}_{\text{water" = "4.18 J g"^(-1)""^@"C}}^{- 1}$

http://www2.ucdsb.on.ca/tiss/stretton/database/specific_heat_capacity_table.html

Now, the specific heat of a substance tells you the amount of heat needed to increase the temperature of $\text{1 g}$ of said substance by ${1}^{\circ} \text{C}$.

In this case, you can say that you need $\text{4.18 J}$ of heat in order to increase the temperature of $\text{1 g}$ of water by ${1}^{\circ} \text{C}$.

Start by calculating the amount of heat needed to increase the temperature of $\text{23 g}$ of water by ${1}^{\circ} \text{C}$

23 color(red)(cancel(color(black)("g"))) * "4.18 J"/(1color(red)(cancel(color(black)("g"))) * 1^@"C") = "96.14 J"""^@"C"^(-1)

This means that in order to increase the temperature of $\text{23 g}$ of water by ${1}^{\circ} \text{C}$, you need to provide it with $\text{96.14 J}$ of heat.

However, you must increase the temperature of the sample by

$\Delta T = {68}^{\circ} \text{C" - 31^@"C" = 37^@"C}$

which implies that you will need to supply the sample with

37 color(red)(cancel(color(black)(""^@"C"))) * "96.14 J"/(1color(red)(cancel(color(black)(""^@"C")))) = "3357.2 J"#

Rounded to two sig figs, the answer will be

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{heat needed = 3400 J}}}}$