# Question f3dcd

Apr 8, 2017

${\text{0.9000 J g"^(-1)""^@"C}}^{- 1}$

#### Explanation:

A substance's specific heat tells you the amount of heat needed to increase the temperature of $\text{1 g}$ of that substance by ${1}^{\circ} \text{C}$.

In your case, you know that it takes $\text{132.8 J}$ of heat to increase the temperature of $\text{11.17 g}$ of aluminium by

$\Delta T = {28.94}^{\circ} \text{C" - 15.73^@"C" = 13.21^@"C}$

The first thing to do here is to figure out how much heat is needed in order to increase the temperature of $\text{11.17 g}$ of aluminium by ${1}^{\circ} \text{C}$.

1 color(red)(cancel(color(black)(""^@"C"))) * overbrace("132.8 J"/(13.21color(red)(cancel(color(black)(""^@"C")))))^(color(blue)("true for 11.17 g of Al")) = "10.053 J"

So, you know that if you deliver $\text{10.053 J}$ of heat to $\text{11.17 g}$ of aluminium, the temperature of the sample will increase by ${1}^{\circ} \text{C}$.

In order to increase the temperature of $\text{1 g}$ of aluminium by ${1}^{\circ} \text{C}$, you will need

1 color(red)(cancel(color(black)("g"))) * overbrace("10.053 J"/(11.17color(red)(cancel(color(black)("g")))))^(color(blue)("true for a 1"^@"C increase in temperature")) = "0.9000 J"#

Therefore, you can say that the specific heat of aluminium is equal to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{c}_{\text{Al" = "0.9000 J g"^(-1)""^@"C}}^{- 1}}}}$

You need to supply $\text{0.9000 J}$ of heat to increase the temperature of $\text{1 g}$ of aluminium by ${1}^{\circ} \text{C}$.

The answer is rounded to four sig figs.

This is an excellent result because aluminium's specific heat is listed as being equal to what we found here.

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