# 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?

##### 1 Answer

#### 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

Now, you know that your sample has a mass of *conversion factor* to find the amount of heat needed to increase the temperature of this much water by

#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 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**.