Question

Question #de7a7

Answer 1

`140^@"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 `"1 g"` of glass by `1^@"C"`.

`c_"glass" = "0.50 J g"^(-1)""^@"C"^(-1)`

So, you know that in order to increase the temperature of `"1 g"` of glass by `1^@"C"`, you need to supply it with `"0.50 J"` of energy.

Now, your sample has a mass of `"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 `"103.2 g"` of glass by `1^@"C"`, you need to provide `"51.6 J"`.

In your case, `"6375 J"` will increase the temperature of `"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^@"C" ~~ 120^@"C"`

Since this represents the increase in temperature that corresponds to a `"103.2-g"` sample of glass that absorbs `"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^@"C"` does not have any decimal places.