Question

Question #9b580

Answer 1

`q = -"9.4 J"`

Explanation:

The first thing to keep in mind here is that when ice cools it actually gives off heat to its surroundings.

This implies that `q` will carry a negative sign since we associate a negative value of `q` with heat being given off.

Now, the specific heat of ice tells you how much heat will be given off when the temperature of `"1 g"` of ice decreases by `1^@"C"` or by `"1 K"`.

More specifically, you now what

`c_"ice" = "2.087 J g"^(-1)"K"^(-1) = "2.087 J g"^(-1)""^@"C"^(-1)`

which means that when the temperature of `"1 g"` of ice decreases by `1^@"C"`, `"2.087 J"` of heat are being given off, i.e.

`q = -"2.087 J" ->` corresponds to a decrease of `1^@"C"` for `"1 g"` of ice

You can start by calculating how much heat will be given off when `"0.12 g"` of ice cools by `1^@"C"`. To do that, use the specific heat of ice as a conversion factor

`0.12 color(red)(cancel(color(black)("g"))) * "2.087 J"/(1color(red)(cancel(color(black)("g"))) * 1^@"C") = "0.25044 J"""^@"C"^(-1)`

This means that `"0.25044 J"` of heat are being given off when `"0.12 g"` of ice is being cooled by `1^@"C"`. In your case, you must cool the sample by

`DeltaT = |-66.0^@"C" - (28.6^@"C")|`

`DeltaT = |-37.4^@"C"| = 37.4^@"C"`

which means that you will have

`37.4 color(red)(cancel(color(black)(""^@"C"))) * overbrace("0.25044 J"/(1color(red)(cancel(color(black)(""^@"C")))))^(color(blue)("for 0.12 g of ice")) = "9.4 J"`

This means that when `"0.12 g"` of ice cools from `-28.6^@"C"` to `-66.0^@"C"`, `"9.4 J"` of heat are being given off, which is equivalent to saying that

`color(darkgreen)(ul(color(black)(q = -"9.4 J")))`

The answer is rounded to two sig figs.

So remember, when heat is being given off, `q` carries a negative sign. When heat is being absorbed, `q` carries a positive sign.