Question #5012c

1 Answer
Mar 11, 2017

Answer:

#"3400 J"#

Explanation:

The first thing you need to do here is to look up the specific heat of water

#c_"water" = "4.18 J g"^(-1)""^@"C"^(-1)#

http://www2.ucdsb.on.ca/tiss/stretton/database/specific_heat_capacity_table.html

Now, the specific heat of a substance tells you the amount of heat needed to increase the temperature of #"1 g"# of said substance by #1^@"C"#.

In this case, you can say that you need #"4.18 J"# of heat in order to increase the temperature of #"1 g"# of water by #1^@"C"#.

Start by calculating the amount of heat needed to increase the temperature of #"23 g"# of water by #1^@"C"#

#23 color(red)(cancel(color(black)("g"))) * "4.18 J"/(1color(red)(cancel(color(black)("g"))) * 1^@"C") = "96.14 J"""^@"C"^(-1)#

This means that in order to increase the temperature of #"23 g"# of water by #1^@"C"#, you need to provide it with #"96.14 J"# of heat.

However, you must increase the temperature of the sample by

#DeltaT = 68^@"C" - 31^@"C" = 37^@"C"#

which implies that you will need to supply the sample with

#37 color(red)(cancel(color(black)(""^@"C"))) * "96.14 J"/(1color(red)(cancel(color(black)(""^@"C")))) = "3357.2 J"#

Rounded to two sig figs, the answer will be

#color(darkgreen)(ul(color(black)("heat needed = 3400 J")))#