# Question 509d0

Mar 24, 2017

The heat required, $Q$, is $\text{4630 J}$.

#### Explanation:

Use the following equation:

$Q = m c \Delta T$,

where $Q$ is the amount of heat energy, $m$ is mass, $c$ is specific heat capacity, $\Delta T$. $\Delta T = {T}_{\text{final"-T_"initial}}$

Given
$m = \text{66.7}$
c="0.75 J/(g"^@"C")
$\Delta T = \text{92.5"^@"C"-0.0^@"C"=92.5^@"C}$

Unknown: $Q$

Solution
Substitute the given values into the equation and solve.

Q=(66.7color(red)cancel(color(black)("g")))xx((0.75color(white)(.)"J")/(color(red)cancel(color(black)("g"^@"C"))))xx(92.5color(red)cancel(color(black)(""^@"C")))="4630 J"

(rounded to three significant figures)

Mar 24, 2017

$\text{4630 J}$

#### Explanation:

Here's an alternative approach that you can use to double-check your calculations.

As you know, a substance's specific heat tells you the amount of heat needed to increase the temperature of $\text{1 g}$ of said substance by ${1}^{\circ} \text{C}$.

In your case, you know that silicon carbide has a specific heat of

${c}_{\text{SiC" = "0.75 J g"^(-1)""^@"C}}^{- 1}$

This tells you that in order to increase the temperature of $\text{1 g}$ of silicon carbide by ${1}^{\circ} \text{C}$, you must provide it with $\text{0.75 J}$ of energy.

Now, you can use the specific heat as a conversion factor to help you determine the amount if heat needed to increase the temperature of $\text{66.7 g}$ of silicon carbide by ${1}^{\circ} \text{C}$

66.7 color(red)(cancel(color(black)("g"))) * "0.75 J"/(1color(red)(cancel(color(black)("g"))) * 1^@"C") = "50.025 J"""^@"C"^(-1)#

Now that you know how much heat will increase the temperature of $\text{66.7 g}$ of silicon carbide by ${1}^{\circ} \text{C}$, you can use this as a conversion factor to find the heat needed to increase the temperature of the sample by

${92.5}^{\circ} \text{C" - 0.0^@"C" = 92.5^@"C}$

You should end up with

$92.5 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{^@"C"))) * overbrace("50.025 J"/(1color(red)(cancel(color(black)(""^@"C")))))^(color(blue)("for 66.7 g SiC")) = color(darkgreen)(ul(color(black)("4630 J}}}}$

The answer is rounded to three sig figs.