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?

A substance's tells you how much heat much either be added or removed from $"1 g"$ of that substance in order to cause a $1^@"C"$ change in temperature.

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

$color(blue)(DeltaT = T_"final" - T_"initial")$

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

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

$color(blue)(q = m * c * DeltaT)" "$, where

$q$ - the amount of heat added / removed
$m$ - the mass of the sample
$c$ - the of the substance
$DeltaT$ - 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 $"5275 J"$ of heat to that $"50.0-g"$ piece of glass will result in a temperature change of

$q = m * c * DeltaT implies DeltaT = q/(m * c)$

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 $211^@"C"$ increase in temperature. This means that the final temperature of the glass will be

$T_"final" = T_"initial" + DeltaT$

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

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