Question

Question #3269e

Answer 1

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")))`