Question

Question #fae70

Answer 1

`"795 kJ"`

Explanation:

You can switch things up a little and solve by converting the specific heat of copper from joules per gram Celsius to kilojoules per kilogram Celsius by using the two conversion factors

`"1 kJ" = 10^3color(white)(.)"J" " "` and `" " "1 kg" = 10^3color(white)(.)"g"`

You will end up with

`0.385 color(white)(.)color(red)(cancel(color(black)("J")))/(1color(red)(cancel(color(black)("g"))) * 1^@"C") * "1 kJ"/(10^3color(red)(cancel(color(black)("J")))) * (10^3color(red)(cancel(color(black)("g"))))/"1 kg" = "0.385 kJ kg"^(-1)""^@"C"^(-1)`

So, you know that in order to increase the temperature of `"1 kg"` of copper by `1^@"C"`, you need to provide it with `"0.385 kJ"` of heat.

You can use the specific heat of copper to figure out the amount of heat needed to increase the temperature of `"6.62 kg"` of copper.

`6.62 color(red)(cancel(color(black)("kg"))) * "0.385 kJ"/(1color(red)(cancel(color(black)("kg"))) * 1^@"C") = "2.549 kJ"""^@"C"^(-1)`

This means that in order to increase the temperature of `"6.62 kg"` of copper by `1^@"C"`, you need to provide the sample with `"2.549 kJ"` of heat.

Finally, you can use this information to determine the amount of heat needed to increase the temperature of `"6.62 kg"` of copper by

`334.3^@"C" - 22.5^@"C" = 311.8^@"C"`

You should get

`311.8color(red)(cancel(color(black)(""^@"C"))) * overbrace("2.549 kJ"/(1color(red)(cancel(color(black)(""^@"C")))))^(color(blue)("for 6.62 kg of copper")) = color(darkgreen)(ul(color(black)("795 kJ")))`

The answer is rounded to three sig figs.