Question

Question #3c48f

Answer 1

Here's what I got.

Explanation:

The key here is the specific heat of water, which is usually given as

`c = "4.18 J g"^(-1)""^@"C"^(-1)`

The specific heat of a substance is the amount of heat that must be provided in order to increase the temperature of `"1 g"` of that substance by `1^@"C"`.

In your case, you need to provide `"4.18 J"` of heat in order to heat `"1 g"` of water by `1^@"C"`.

Now, the problem provides you the amount of heat that you have at your disposal and the desired change in temperature, which is equal to

`DeltaT = 75.0^@"C" - 20.0^@"C" = 55.0^@"C"`

Use the specific heat of water to calculate how much heat would be needed in order to increase the temperature of `"1 g"` of water by `55^@"C"`

`55.0 color(red)(cancel(color(black)(""^@"C"))) * "4.18 J"/("1 g" * 1color(red)(cancel(color(black)(""^@"C")))) = "229.9 J g"^(-1)`

So, you must add `"229.9 J"` for every `"1 g"` of water in order to cause its temperature to increase by `55.0^@"C"`. This means that `"8000 J"` of heat will get you

`8000 color(red)(cancel(color(black)("J"))) * overbrace("1 g"/(229.9 color(red)(cancel(color(black)("J")))))^(color(blue)("needed for 55" ""^@"C increase")) = color(darkgreen)(ul(color(black)("35.8 g")))`

I'll leave the answer rounded to three sig figs, but keep in mind that you only provided one significant figure for the amount of heat you have at your disposal.