Question #4113d

1 Answer
Aug 21, 2017

Answer:

#"9,500 J"#

Explanation:

An important assumption to make here is that you're increasing the temperature of liquid water by #87^@"C"#. In other words, you should assume that this increase in temperature does not include a phase change.

Now, the key here is water's specific heat

#c_"water" = "4.18 J g"^(-1)""^@"C"^(-1)#

This tells you the amount of energy needed to increase the temperature of #"1 g"# of water by #1^@"C"#.

So, for a sample of #"1-g"# of water, you need #"4.18 J"# of heat to increase its temperature by #1^@"C"#. This means that for any sample of water, you will need

#87 color(red)(cancel(color(black)(""^@"C"))) * overbrace("4.18 J"/("1 g" * 1color(red)(cancel(color(black)(""^@"C")))))^(color(blue)("water's specific heat")) = "363.7 J g"^(-1)#

to increase its temperature by #87^@"C"#. Since your sample has a mass of #"26 g"#, it follows that you will need a total of

#26 color(red)(cancel(color(black)("g"))) * overbrace("363.7 J"/(1color(red)(cancel(color(black)("g")))))^(color(blue)("for a 87-"^@"C increase")) = color(darkgreen)(ul(color(black)("9,500 J")))#

The answer is rounded to two sig figs.