Question

Question #779cf

Answer 1

`"19,500 J"`

Explanation:

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

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

Now, the specific heat of a given substance tells you the amount of heat needed to increase the temperature of `"1 g"` of said substance by `1^@"C"`.

In the case of water, you have

`c_"water" = color(blue)("4.18 J") color(white)(.)color(red)("g"^(-1) color(white)(.)color(purple)(""^@"C"^(-1))`

which means that you need `color(blue)("4.18 J")` of heat to raise the temperature of `color(red)("1 g")` of water by `1^@"C"`.

Now, you need to raise the temperature of `"85.0 g"` of water by

`DeltaT = 80.0^@"C" - 25.0^@"C" = 55.0^@"C"`

The first thing to do here is to calculate how much heat will be needed to raise the temperature of `"85.0 g"` of water by `1^@"C"`. To do that, use the specific heat of water as a conversion factor

`85.0 color(red)(cancel(color(black)("g"))) * "4.18 J"/(1color(red)(cancel(color(black)("g"))) * 1^@"C") = "355.3 J" ""^@"C"^(-1)`

This means that for every `1^@"C"` increase in the temperature of an `"85.0-g"` sample of water, you need `"355.3 J"` of heat.

In your case, you will need

`55.0 color(red)(cancel(color(black)(""^@"C"))) * "355.3 J"/(1color(red)(cancel(color(black)(""^@"C")))) = color(darkgreen)(ul(color(black)("19,500 J")))`

The answer is rounded to three sig figs.

You can double-check this result by using the equation

`color(blue)(ul(color(black)(q = m * c * DeltaT)))`

Here

• `q` is the heat lost or gained by the substance
• `m` is the mass of the sample
• `c` is the specific heat of the substance
• `DeltaT` is the change in temperature, defined as the difference between the final temperature and the initial temperature of the sample

In your case, you will have

`q = 85.0 color(red)(cancel(color(black)("g"))) * "4.18 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * 55.0 color(red)(cancel(color(black)(""^@"C")))`

`q = color(darkgreen)(ul(color(black)("19,500 J")))`