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
Jan 11, 2017

Answer:

#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.