Question #3269e

1 Answer
Jul 8, 2017

Answer:

Here's what I got.

Explanation:

I'm going to assume that you're interested in determining the amount of heat needed to convert #"89.5 g"# of liquid water at its normal melting point of #0^@"C"# to liquid water at its normal boiling point of #100^@"C"#.

In other words, I will assume that you don't have to go from solid water at #0^@"C"# to water vapor at #100^@"C"#, i.e. that no phase change is involved here.

Now, the specific heat of water, which tells you the amount of heat needed to increase the temperature of #"1 g"# of water by #1^@"C"#, is equal to #"4.18 J g"^(-1)""^@"C"^(-1)#. This means that in order to increase the temperature of #"1 g"# of liquid water by #1^@"C"#, you need to provide it with #"4.18 J"# of heat.

In your case, you would need

#89.5 color(red)(cancel(color(black)("g"))) * "4.18 J"/(1color(red)(cancel(color(black)("g"))) * 1^@"C") = "374.11 J"""^@"C"^(-1)#

in order to increase the temperature of your sample of water. This value tells you that every #1^@"C"# increase in the temperature of your sample requires #"374.11 J"# of heat.

You can thus say that a change in temperature of

#100^@"C" - 0^@"C" = 100^@"C"#

will require

#100 color(red)(cancel(color(black)(""^@"C"))) * overbrace("374.11 J"/(1color(red)(cancel(color(black)(""^@"C")))))^(color(blue)("for 89.5 g of water")) = "37,411 J"#

of heat. Rounded to three sig figs, the number of significant figures you have for the mass of water, and expressed in kilojoules, the answer will be

#color(darkgreen)(ul(color(black)("heat needed = 37.4 kJ")))#