# A 0.050g ball of aluminum foil absorbs 1.42 J heat when it is heated 31.6°C. What is the specific heat of aluminum foil?

Aug 13, 2016

${\text{0.899 J g"^(-1)""^@"C}}^{- 1}$

#### 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 the temperature of $\text{0.050 g}$ of aluminium increased by ${31.6}^{\circ} \text{C}$ upon the absorption of $\text{1.42 J}$ of heat.

The first thing to do here is to figure out how much heat is needed to increase the temperature of $\text{1 g}$ of aluminium by ${31.6}^{\circ} \text{C}$.

1 color(red)(cancel(color(black)("g"))) * "1.41 J"/(0.050color(red)(cancel(color(black)("g")))) = "28.4 J"

This is how much heat is needed to cause a ${31.6}^{\circ} \text{C}$ increase in the temperature of $\text{1 g}$ of aluminium. Now all you have to do is calculate how much heat is needed to increase the temperature of $\text{1 g}$ by ${1}^{\circ} \text{C}$

1 color(red)(cancel(color(black)(""^@"C"))) * "28.4 J"/(31.6color(red)(cancel(color(black)(""^@"C")))) = "0.899 J"

Therefore, the specific heat of aluminium will be

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{c}_{\text{Al" = "0.899 J g"^(-1)""^@"C}}^{- 1}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to three sig figs.

$\textcolor{w h i t e}{\frac{a}{a}}$

ALTERNATIVELY

You can get the same result by suing the following equation

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} q = m \cdot c \cdot \Delta T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, where

$q$ - the amount of heat released
$m$ - the mass of the sample
$c$ - the specific heat of the substance
$\Delta T$ - the change in temperature, defined as the difference between the final temperature and the initial temperature

In your case, the change in temperature is equal to

$\Delta T = {31.6}^{\circ} \text{C}$

You will have

$q = m \cdot c \cdot \Delta T \implies c = \frac{q}{m \cdot \Delta T}$

Plug in your values to find

c_"Al" = "1.42 J"/("0.050 g" * 31.6^@"C") = color(green)(|bar(ul(color(white)(a/a)color(black)("0.899 J g"^(-1)""^@"C"^(-1))color(white)(a/a)|)))

The listed value for the specific heat of aluminium is

${c}_{\text{Al" = "0.900 J g"^(-1)""^@"C}}^{- 1}$

http://hyperphysics.phy-astr.gsu.edu/hbase/tables/sphtt.html