# Question a6d1a

Dec 17, 2015

$2.4 \cdot {10}^{6} \text{J}$

#### Explanation:

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

In your case, graphite is said to have a specific heat equal to 0.71"J"/("g" ""^@"C")#. Now, before doing anything else, take a second to analyze the units for the specific heat of graphite and the units given to you in the problem.

First thing to notice here is that you are given a mass of graphite expressed in kilograms, but that the specific heat values uses grams. So a unit conversion is needed here.

Likewise, the temperature increase is given to you in Kelvin, but the specific heat value uses degrees Celsius.

Now, when it comes to increase or decrease in temperature, it makes no difference which temperature scale is used. For example, the change in temperature will be

$\Delta T = \text{348 K" - "294 K" = "54 K}$

Calculated in degrees Celsius, you have

$\Delta T = {\left(348 - 273.15\right)}^{\circ} \text{C" - (294 - 273.15)^@"C}$

$\Delta T = {348}^{\circ} \text{C" - color(red)(cancel(color(black)(273.15^@"C"))) - 294^@"C" + color(red)(cancel(color(black)(273.15^@"C"))) = 54^@"C}$

This means that you can use the change in temperature interchangeably for Kelvin and degrees Celsius.

Now, the equation that establishes a relationship between heat absorbed/lost and change in temperature looks like this

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

$q$ - heat absorbed/lost
$m$ - 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

Plug in your values and solve for $q$ - do not forget about converting the mass of graphite from kilograms to grams

$q = 63 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{kg"))) * (1000color(red)(cancel(color(black)("g"))))/(1color(red)(cancel(color(black)("kg")))) * 0.71"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * 54color(red)(cancel(color(black)(""^@"C}}}}$

$q = \text{2,415,420 J}$

Rounded to two sig figs, the answer will be

$q = \textcolor{g r e e n}{2.4 \cdot {10}^{6} \text{J}}$