# Question 44228

Dec 2, 2015

$\text{728 J}$

#### Explanation:

All you have to do here is use the equation the establishes a relationship, in your case, between heat gained and increase in temperature

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

$q$ - heat absorbed
$m$ - the mass of the sample
$c$ - the specific heat of the substance
$\Delta T$ - the change in temperature, defined as final temperature minus initial temperature

So, you know that copper has a specific heat of

c_"copper" = 0.385"J"/("g" ""^@"C")

So, what does a substance's specific heat tell you?

Well, it tells you how much heat is needed to increase the mass of a $\text{1.00-g}$ sample by ${1}^{\circ} \text{C}$. More specifically, you need to provide $\text{0.385 J}$ of heat to increase the mass of $\text{1.00 g}$ of copper by ${1}^{\circ} \text{C}$.

Now, if you have a bigger mass, you'd need more heat to increase its temperature by ${1}^{\circ} \text{C}$.

If you also want to increase its temperature by more than ${1}^{\circ} \text{C}$, you'd once again need more heat.

In your case, the change in temperature will be

$\Delta T = {324.3}^{\circ} \text{C" - 20.5^@"C" = 303.8^@"C}$

This means that the amount of heat you'd need will be equal to

q = 6.22 color(red)(cancel(color(black)("g"))) * 0.385"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * 303.8color(red)(cancel(color(black)(""^@"C"))) = "727.51 J"#

Rounded to three sig figs, the answer will be

$q = \textcolor{g r e e n}{\text{728 J}}$