Question #9b580

1 Answer
Jan 30, 2017

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.