# Question 6493b

Sep 15, 2016

$\text{0.65 kcal}$

#### Explanation:

The first thing to do here is to make sure that you have a clear understanding of what a substance's specific heat tells you.

Specific heat is defined as the amount of heat needed to raise 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 ${\text{0.90 J g"^(-1)""^@"C}}^{- 1}$. This means that if you start with $\text{1.0 g}$ of aluminium, you can raise its temperature by ${1}^{\circ} \text{C}$ by adding $\text{0.90 J}$ of heat.

Now, your sample contains more than $\text{1 g}$ of aluminium. The first thing to do here is to figure out how much heat is needed to increase the temperature of $\text{78 g}$ of aluminium by ${1}^{\circ} \text{C}$.

To do that, use the specific heat!

78 color(red)(cancel(color(black)("g"))) * overbrace("0.90 J"/(1color(red)(cancel(color(black)("g"))) ""^@"C"))^(color(darkgreen)("the specific heat of aluminium")) = "70.2 J"""^@"C"^(-1)

This means that in order to increase the temperature of $\text{78 g}$ of aluminium by ${1}^{\circ} \text{C}$, you must provide $\text{70.2 J}$ of heat. How about if you want to increase its temperature by

$\text{temperature increase" = 89^@"C" - 50^@"C" = 39^@"C}$

how much heat would be needed in this case?

39 color(red)(cancel(color(black)(""^@"C"))) * overbrace("70.2 J"/(1color(red)(cancel(color(black)(""^@"C")))))^(color(darkgreen)("for 78 g of aluminium")) = "2737.8 J"

Finally, you must convert this to kilocalories by using

color(purple)(bar(ul(|color(white)(a/a)color(black)("1 cal " = " 4.18 J")color(white)(a/a)|)))" " and " "color(purple)(bar(ul(|color(white)(a/a)color(black)("1 kcal " = 10^3"cal")color(white)(a/a)|)))

You will have

2737.8 color(red)(cancel(color(black)("J"))) * (1color(red)(cancel(color(black)("cal"))))/(4.18color(red)(cancel(color(black)("J")))) * "1 kcal"/(10^3color(red)(cancel(color(black)("cal")))) = color(green)(bar(ul(|color(white)(a/a)color(black)("0.65 kcal")color(white)(a/a)|)))#

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one sig fig for the initial temperature of the sample, i.e. ${50}^{\circ} \text{C}$.