Question #44228

1 Answer
Dec 2, 2015

"728 J"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)" "q=mcΔT , where

qq - heat absorbed
mm - the mass of the sample
cc - the specific heat of the substance
DeltaTΔT - 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")ccopper=0.385JgC

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"1.00-g sample by 1^@"C"1C. More specifically, you need to provide "0.385 J"0.385 J of heat to increase the mass of "1.00 g"1.00 g of copper by 1^@"C"1C.

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

If you also want to increase its temperature by more than 1^@"C"1C, 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"ΔT=324.3C20.5C=303.8C

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")