# Question #9b580

##### 1 Answer

#### 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 *negative sign* since we associate a negative value of

Now, the **specific heat** of ice tells you how much heat will be given off when the temperature of *decreases* by **or** by

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 **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 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

#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 **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**, **absorbed**,