# Question #5012c

##### 1 Answer

#### 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

In this case, you can say that you need

Start by calculating the amount of heat needed to increase the temperature of

#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

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")))#