# How many joules of heat are needed to raise the temperature of 10.0 g of aluminum from 22°C to 55°C, if the specific heat of aluminum is 0.90 J/g°C?

Jan 28, 2016

$\text{297 J}$

#### Explanation:

The key to this problem lies with aluminium's specific heat, which as you know tells you how much heat is needed in order to increase the temperature of $\text{1 g}$ of a given substance by ${1}^{\circ} \text{C}$.

In your case, aluminium is said to have a specific heat of 0.90 "J"/("g" ""^@"C").

So, what does that tell you?

In order to increase the temperature of $\text{1 g}$ of aluminium by ${1}^{\circ} \text{C}$, you need to provide it with $\text{0.90 J}$ of heat.

But remember, this is how much you need to provide for every gram of aluminium in order to increase its temperature by ${1}^{\circ} \text{C}$. So if you wanted to increase the temperature of $\text{10.0 g}$ of aluminium by ${1}^{\circ} \text{C}$, you'd have to provide it with

underbrace(overbrace("0.90 J")^(color(blue)("1 gram")) + overbrace("0.90 J")^(color(blue)("1 gram")) + " ... " + overbrace("0.90 J")^(color(blue)("1 gram")))_(color(red)("10 times")) = 10 xx "0.90 J"

However, you don't want to increase the temperature of the sample by ${1}^{\circ} \text{C}$, you want to increase it by

$\Delta T = {55}^{\circ} \text{C" - 22^@"C" = 33^@"C}$

This means that you're going to have to use that much heat for every degree Celsius you want the temperature to change. You can thus say that

underbrace(overbrace(10 xx "0.90 J")^(color(purple)(1^@"C")) + overbrace(10 xx "0.90 J")^(color(purple)(1^@"C")) + " ... " + overbrace(10 xx "0.90 J")^(color(purple)(1^@"C")))_(color(red)("33 times")) = 33 xx 10 xx "0.90 J"

Therefore, the total amount of heat needed to increase the temperature of $\text{10.0 g}$ of aluminium by ${33}^{\circ} \text{C}$ will be

$q = 10.0 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * 0.90"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * 33color(red)(cancel(color(black)(""^@"C}}}}$

$q = \textcolor{g r e e n}{\text{297 J}}$

I'll leave the answer rounded to three sig figs, despite the fact that your values only justify two sig figs.

For future reference, this equation will come in handy

$\textcolor{b l u e}{q = m \cdot c \cdot \Delta T} \text{ }$, where

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