Question

A 36-gram sample of water has an initial temperature of 22°C. After the sample absorbs 1200 joules of heat energy, what is the final temperature of the sample?

Answer 1

`30.^@"C"`

Explanation:

The key here is the specific heat of water, which you'll usually find list as

`c = "4.18 J g"^(-1)""^@"C"^(-1)`

The specific heat of a substance tells you how much heat is needed in order to increase the temperature of `"1 g"` of that substance by `1^@"C"`. In water's case, it takes `"4.18 J"` of heat to increase the temperature of `"1 g"` of water by `1^@"C"`.

Now, you know that your sample has a mass of `"36 g"` and that you supply it with `"1200 J"` of heat. Use the specific heat of water as a conversion factor to find the amount of heat needed to increase the temperature of this much water by `1^@"C"`

`36 color(red)(cancel(color(black)("g"))) * "4.18 J"/(1color(red)(cancel(color(black)("g"))) * 1^@"C") = "150.5 J"""^@"C"^(-1)`

This means that `"1200 J"` of heat will increase the temperature of the sample by

`1200 color(red)(cancel(color(black)("J"))) * (1^@"C")/(150.5color(red)(cancel(color(black)("J")))) = 7.973^@"C"`

Therefore, the final temperature of the sample will be

`T_"final" = 22^@"C" + 7.973^@"C" = color(darkgreen)(ul(color(black)(30.^@"C")))`

I'll leave the answer rounded to two sig figs.