Question #44228

1 Answer
Dec 2, 2015

Answer:

#"728 J"#

Explanation:

All you have to do here is use the equation the establishes a relationship, in your case, between heat gained and increase in temperature

#color(blue)(q = m * c * DeltaT)" "#, where

#q# - heat absorbed
#m# - the mass of the sample
#c# - the specific heat of the substance
#DeltaT# - the change in temperature, defined as final temperature minus initial temperature

So, you know that copper has a specific heat of

#c_"copper" = 0.385"J"/("g" ""^@"C")#

So, what does a substance's specific heat tell you?

Well, it tells you how much heat is needed to increase the mass of a #"1.00-g"# sample by #1^@"C"#. More specifically, you need to provide #"0.385 J"# of heat to increase the mass of #"1.00 g"# of copper by #1^@"C"#.

Now, if you have a bigger mass, you'd need more heat to increase its temperature by #1^@"C"#.

If you also want to increase its temperature by more than #1^@"C"#, you'd once again need more heat.

In your case, the change in temperature will be

#DeltaT = 324.3^@"C" - 20.5^@"C" = 303.8^@"C"#

This means that the amount of heat you'd need will be equal to

#q = 6.22 color(red)(cancel(color(black)("g"))) * 0.385"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * 303.8color(red)(cancel(color(black)(""^@"C"))) = "727.51 J"#

Rounded to three sig figs, the answer will be

#q = color(green)("728 J")#