Question #fae70

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.