Question #9b580

1 Answer
Jan 30, 2017

Answer:

#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.