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
Now, the specific heat of ice tells you how much heat will be given off when the temperature of
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
q = -"2.087 J" -> corresponds to a decrease of1^@"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
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,