Question

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?

Answer 1

`230^@"C"`

Explanation:

A substance's specific heat 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 specific heat 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 sig figs, the number of sig figs you have for the specific heat of glass.