# Question de7a7

Sep 17, 2017

${140}^{\circ} \text{C}$

#### Explanation:

The key here is the specific heat of glass because it tells you the amount of heat needed in order to increase the mass of $\text{1 g}$ of glass by ${1}^{\circ} \text{C}$.

${c}_{\text{glass" = "0.50 J g"^(-1)""^@"C}}^{- 1}$

So, you know that in order to increase the temperature of $\text{1 g}$ of glass by ${1}^{\circ} \text{C}$, you need to supply it with $\text{0.50 J}$ of energy.

Now, your sample has a mass of $\text{103.2 g}$. Use the specific heat of glass to calculate how much heat is needed to increase the temperature of this sample

103.2 color(red)(cancel(color(black)("g"))) * overbrace("0.50 J"/(1 color(red)(cancel(color(black)("g"))) * 1^@"C"))^(color(blue)("the specific heat of glass")) = "51.6 J"""^@"C"^(-1)

This tells you that in order to increase the temperature of $\text{103.2 g}$ of glass by ${1}^{\circ} \text{C}$, you need to provide $\text{51.6 J}$.

In your case, $\text{6375 J}$ will increase the temperature of $\text{103.2 g}$ of glass by

6375 color(red)(cancel(color(black)("J"))) * (1^@"C")/(51.6 color(red)(cancel(color(black)("J")))) = 123.55^@"C"

At this point, you should round the value to two sig figs because that's how many sig figs you have for the specific heat of glass.

So you will have

${123.55}^{\circ} \text{C" ~~ 120^@"C}$

Since this represents the increase in temperature that corresponds to a $\text{103.2-g}$ sample of glass that absorbs $\text{6375 J}$, you can say that the final temperature of the glass will be

T_"final" = 20.0^@"C" + 120^@"C" = color(darkgreen)(ul(color(black)(140^@"C")))#

The answer cannot have any decimal places because ${120}^{\circ} \text{C}$ does not have any decimal places.