Question

Question #5012c

Answer 1

`"3400 J"`

Explanation:

The first thing you need to do here is to look up the specific heat of water

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

http://www2.ucdsb.on.ca/tiss/stretton/database/specific_heat_capacity_table.html

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

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

Start by calculating the amount of heat needed to increase the temperature of `"23 g"` of water by `1^@"C"`

`23 color(red)(cancel(color(black)("g"))) * "4.18 J"/(1color(red)(cancel(color(black)("g"))) * 1^@"C") = "96.14 J"""^@"C"^(-1)`

This means that in order to increase the temperature of `"23 g"` of water by `1^@"C"`, you need to provide it with `"96.14 J"` of heat.

However, you must increase the temperature of the sample by

`DeltaT = 68^@"C" - 31^@"C" = 37^@"C"`

which implies that you will need to supply the sample with

`37 color(red)(cancel(color(black)(""^@"C"))) * "96.14 J"/(1color(red)(cancel(color(black)(""^@"C")))) = "3357.2 J"`

Rounded to two sig figs, the answer will be

`color(darkgreen)(ul(color(black)("heat needed = 3400 J")))`