Question

Question #4113d

Answer 1

`"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.