# What quantity of heat energy must have been applied to a block of aluminum weighing 42.7 g if the temperature of the block of aluminum increased by 15.2°C?

Mar 11, 2016

$\text{591 J}$

#### Explanation:

As it stands, the problem fails to provide you with an essential piece of information - the value of aluminium's specific heat.

This means that you're going to have to track it yourself. You can find it listed here

http://www.engineeringtoolbox.com/specific-heat-metals-d_152.html

as being equal to

$c = {\text{0.91 J g"^(-1)""^@"C}}^{- 1}$

So, a substance's specific heat tells you how much heat is needed in order to increase the temperature of $\text{1 g}$ of that substance by ${1}^{\circ} \text{C}$.

In this case, you need to provide aluminium with $\text{0.91 J}$ of heat in order to increase the temperature of $\text{1 g}$ by ${1}^{\circ} \text{C}$.

You know that the block of aluminium has a mass $\text{42.7 g}$. This means that if you were to increase its temperature by just ${1}^{\circ} \text{C}$, you'd need

42.7 color(red)(cancel(color(black)("g"))) xx "0.91 J" color(red)(cancel(color(black)("g"^(-1))))""^@"C"^(-1) = "38.86 J" ""^@"C"^(-1)

This much heat would increase the temperature of $\text{42.7 g}$ of aluminium by ${1}^{\circ} \text{C}$. To increase its temperature by ${15.2}^{\circ} \text{C}$, you'd need $15.2$ times more heat

15.2color(red)(cancel(color(black)(""^@"C"))) xx "38.86 J" color(red)(cancel(color(black)(""^@"C"^(-1)))) = "590.67 J"

Rounded to three sig figs, the answer will be

"heat needed" = color(green)(|bar(ul(color(white)(a/a)"591 J"color(white)(a/a)|)))

Alternatively, you can use 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 / removed
$m$ - the mass of the sample
$c$ - the specific heat of the substance
$\Delta T$ - the change in temperature

Plug in your values to get

$q = 42.7 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * "0.91 J" color(red)(cancel(color(black)("g"^(-1))))color(red)(cancel(color(black)(""^@"C"^(-1)))) * 15.2color(red)(cancel(color(black)(""^@"C}}}}$

$q = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{591 J} \textcolor{w h i t e}{\frac{a}{a}} |}}}$