Question

Question #71e95

Answer 1

`58.2^@"C"`

Explanation:

In order to be able to solve this problem, you need to know the value of water's specific heat, which you'll find listed as

`color(purple)(|bar(ul(color(white)(a/a)color(black)(c_"water" = "4.18 J g"^(-1)""^@"C"^(-1))color(white)(a/a)|)))`

A substance's specific heat tells you how much heat is needed in order to increase the temperature of `"1 g"` of that substance by `1^@"C"`.

In your case, you can say that in order to increase the temperature of `"1 g"` of water by `1^@"C"`, you need to provide it with `"4.18 J"` of heat.

Keep in mind that this much heat must be applied per gram, per degree Celsius.

Now, you know that your sample of water has a mass of `"150. g"`, and that it absorbs a total of `"21.8 kJ"` of heat. Convert the heat from kilojoules to joules by using the conversion factor

`"1 kJ" = 10^3"J"`

to get

`21.8color(red)(cancel(color(black)("kJ"))) * (10^3"J")/(1color(red)(cancel(color(black)("kJ")))) = "21,800 J"`

At this point, you can determine how much heat would be needed in order to increase the temperature of this sample of water by `1^@"C"`. Well, you know that you need `"4.18 J"` per gram in order to get a `1^@"C"` increase in temperature, so `"150. g"` of water will require

`150. color(red)(cancel(color(black)("g"))) * overbrace("4.18 J"/(1color(red)(cancel(color(black)("g")))))^(color(purple)("heat needed for 1"^@"C increase for 1 g")) = "627 J"`

If you need `"627 J"` of heat in order to increase the temperature of `"150 g"` of water by `1^@"C"`, it follows that the difference between this much heat and the `"21,800 J"` you supplied went in increasing the sample's temperature.

You can thus say that

`"21,800"color(red)(cancel(color(black)("J"))) * overbrace((1^@"C")/(627color(red)(cancel(color(black)("J")))))^(color(blue)("heat needed for 1"^@"C for 150. g")) = 34.77^@"C"`

If the sample started at an initial temperature of `23.4^@"C"`, you can say that its final temperature will be

`T_"final" = 23.4^@"C" + 34.77^@"C" = color(green)(|bar(ul(color(white)(a/a)58.2^@"C"color(white)(a/a)|)))`

The answer is rounded to three sig figs.

ALTERNATIVE APPROACH

You can get the same result by suing the equation

`color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))" "`, where

`q` - the amount of heat gained / lost
`m` - the mass of the sample
`c` - the specific heat of the substance
`DeltaT` - the change in temperature, defined as the difference between the final temperature and the initial temperature

In your case, you'd have

`DeltaT = q/(m * c)`

`DeltaT = ("21,800" color(red)(cancel(color(black)("J"))))/(150. color(red)(cancel(color(black)("g"))) * 4.18color(red)(cancel(color(black)("J")))color(red)(cancel(color(black)("g"^(-1))))""^@"C"^(-1)) = 34.77^@"C"`

Once again, you'd have

`T_"final" = 23.4^@"C" + DeltaT = color(green)(|bar(ul(color(white)(a/a)58.2^@"C"color(white)(a/a)|)))`