Question #21fd7

The key here is the specific heat of copper, which tells you the amount of energy needed to increase the temperature of #"1 g"# of copper by #1^@"C"#.

#c_"Cu" = "0.385 J g"^(-1)""^@"C"^(-1)#

This tells you that in order to increase the temperature of #"1 g"# of copper by #1^@"C"#, you need to provide the sample with #"0.385 J"# of heat.

Now, you know that you supplied a total of #"280.0 J"# of heat to your sample and that its temperature increased by #26.4^@"C"#.

Use the specific heat of the sample to find the heat needed to increase the temperature of copper by #26.4^@"C"#.

#26.4 color(red)(cancel(color(black)(""^@"C"))) * overbrace("0.385 J"/("1 g" * 1 color(red)(cancel(color(black)(""^@"C")))))^(color(blue)("the specific heat of copper")) = "10.164 J g"^(-1)#

This tells you that in order to increase the temperature of #"1 g"# of copper by #26.4^@"C"#, you need #"10.164 J"# of heat.

You can thus say that the mass of the sample is equal to

#280.0 color(red)(cancel(color(black)("J"))) * overbrace("1 g"/(10.164color(red)(cancel(color(black)("J")))))^(color(blue)("for a 26.4-"""^@"C increase")) = "27.55 g"#

Finally, to convert this to moles, use the molar mass of copper

#27.55 color(red)(cancel(color(black)("g"))) * "1 mole Cu"/(63.546color(red)(cancel(color(black)("g")))) = color(darkgreen)(ul(color(black)("0.434 moles Cu")))#

The answer is rounded to three sig figs.