# Question 5dc43

Mar 20, 2016

$\text{0.015 kg Al}$

#### Explanation:

The key to this problem is aluminium's specific heat, which tells you how much heat is needed to increase the temperature of $\text{1 g}$ of aluminium by ${1}^{\circ} \text{C}$.

Aluminium's specific heat is said to be equal to ${\text{0.897 J g"^(-1)""^@"C}}^{- 1}$. This tells you that in order to increase the temperature of $\text{1 g}$ of aluminium by ${1}^{\circ} \text{C}$, you need to supply it with $\text{0.897 J}$ of heat.

Now, let's assume that your sample has a mass of $\text{1 g}$. In order to increase its temperature by

$\Delta T = {187}^{\circ} \text{C" - 65^@"C" = 122^@"C}$

you would need to provide it with $122$ times more heat than what would be needed to increase its temperature by ${1}^{\circ} \text{C}$.

More specifically, you would need to provide it with

122color(red)(cancel(color(black)(""^@"C"))) * overbrace("0.897 J"/(1color(red)(cancel(color(black)(""^@"C")))))^(color(purple)("heat needed for"color(white)(a)1^@"C"color(white)(a) "increase for 1 g of Al")) = "109.434 J"

However, you end up adding $\text{1650 J}$ of heat to this sample, which of course means that its mass is not equal to $\text{1 g}$.

The ratio between the heat added to the sample to get its temperature to increase by ${122}^{\circ} \text{C}$ and the heat that would be needed to increase the temperature of $\text{1 g}$ of aluminium by ${122}^{\circ} \text{C}$ will give you the mass of the sample, expressed in grams

1650color(red)(cancel(color(black)("J"))) * overbrace("1 g"/(109.434color(red)(cancel(color(black)("J")))))^(color(purple)("heat needed for"color(white)(a) 122^@"C" color(white)(a)"increase for 1 g of Al")) = "15.08 g"

Expressed in kilograms and rounded to two sig figs, the answer will be

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{1 kg" 1= 10^3"g}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

"mass of Al" = color(green)(|bar(ul(color(white)(a/a)"0.015 kg"color(white)(a/a)|)))

ALTERNATIVE SOLUTION

You can get the same result by using the formula

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

$q$ - the amount of heat added
$m$ - the mass of the sample
$c$ - the specific heat of the substance
$\Delta T$ - the change in temperature

Rearrange to solve for $m$

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

Plug in your values to get

$m = \left(1650 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{J"))))/(0.897color(red)(cancel(color(black)("J"))) "g" color(red)(cancel(color(black)(""^@"C"^(-1)))) * (187 - 65)color(red)(cancel(color(black)(""^@"C}}}}\right)$

$m = \text{15.08 g}$

Once again, you'll have

"mass of Al" = color(green)(|bar(ul(color(white)(a/a)"0.015 kg"color(white)(a/a)|)))#