Question

Question #44228

Answer 1

`"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")`