Question 6ad9d

Dec 11, 2016

$\text{2250 J}$

Explanation:

The key to this problem is the specific heat of aluminium

c_"Al" = color(purple)("0.899 J")/(color(green)("g")color(blue)(""^@"C"))

The specific heat of a substance tells you how much heat is required to reaise the temperature of $\text{1 g}$ of said substance by ${1}^{\circ} \text{C}$.

In your case, you know that you need color(purple)("0.899 J" of heat in order to increasde the temperature of color(green)("1 g" of aluminium by $\textcolor{b l u e}{{1}^{\circ} \text{C}}$.

You can use the specific heat of aluminium to calculate how much heat would be required to raise the temperature of $\text{500.0 g}$ of aluminium by ${1}^{\circ} \text{C}$

500.0 color(red)(cancel(color(black)("g"))) * color(purple)("0.899 J")/(color(green)(1)color(red)(cancel(color(green)("g"))) color(blue)(1^@"C")) = "449.5 J"color(blue)(""^@"C"^(-1))

This tells you that you need to provide $\text{449.5 J}$ in order to raise the heat of $\text{500.0 g}$ of aluminium by ${1}^{\circ} \text{C}$. The problem tells you that the temperature of the sample must increase by

$\Delta T = {20.0}^{\circ} \text{C" - 15.0^@"C" = 5.0^@"C}$

In order to get thatto happen, you must provide it with

5.0 color(red)(cancel(color(black)(""^@"C"))) * "449.5 J"/(color(blue)(1)color(red)(cancel(color(blue)(""^@"C")))) = ul("2250 J")#

The answer is rounded to three sig figs.

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

you can also use the following equation

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{q = m \cdot c \cdot \Delta T}}}$

Here

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

In your case, you will once again end up with

$q = 500.0 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * "0.899 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * (20.0 - 15.0) color(red)(cancel(color(black)(""^@"C}}}}$

$q = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{2250 J}}}}$