# To what temperature will a 50.0 g piece of glass raise if it absorbs 5275 joules of heat and its specific heat capacity is 0.50 J/g°C, if the initial temperature of the glass is 20.0°C?

Jan 29, 2016

#### Answer:

${230}^{\circ} \text{C}$

#### Explanation:

A substance's specific heat tells you how much heat much either be added or removed from $\text{1 g}$ of that substance in order to cause a ${1}^{\circ} \text{C}$ change in temperature.

The change in temperature, $\Delta T$, is always calculated by subtracting the initial temperature of the sample from the final temperature of the sample.

$\textcolor{b l u e}{\Delta T = {T}_{\text{final" - T_"initial}}}$

Now, when the substance absorbs heat, its temperature will increase, which implies that $\Delta T > 0$.

Your goal here will be to find the change in temperature first, then use it to find the final temperature of the sample.

You will have to use this equation

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

$q$ - the amount of heat added / removed
$m$ - the mass of the sample
$c$ - the specific heat of the substance
$\Delta T$ - the change in temperature

As you can see, this equation establishes a relationship between the amount of heat added or removed from a sample, the mass of that substance, its specific heat, and the resulting change in temperature.

In your case, adding $\text{5275 J}$ of heat to that $\text{50.0-g}$ piece of glass will result in a temperature change of

$q = m \cdot c \cdot \Delta T \implies \Delta T = \frac{q}{m \cdot c}$

Plug in your values to get

DeltaT = (5275 color(red)(cancel(color(black)("J"))))/(50.0color(red)(cancel(color(black)("g"))) * 0.50color(red)(cancel(color(black)("J")))/(color(red)(cancel(color(black)("g"))) ""^@"C")) = 211^@"C"

So, adding that much heat to your sample will result in a ${211}^{\circ} \text{C}$ increase in temperature. This means that the final temperature of the glass will be

${T}_{\text{final" = T_"initial}} + \Delta T$

T_"final" = 20.0^@"C" + 211^@"C" = color(green)(230^@"C")

The answer is rounded to two sig figs, the number of sig figs you have for the specific heat of glass.