Question

Question #b29dc

Answer 1

`DeltaT = 38^@"C"`

Explanation:

The key to understanding this type of problems lies with the substance's specific heat.

For a given substance, its specific heat will tell you how much heat is needed in order to increase the mass of `"1 g"` of that substance by `1^@"C"`.

In your case, the specific heat of gold is said to be

`c = 0.128"J"/("g" ""^@"C")`

This tells you that you need `"0.128 J"` of heat for every gram of the sample and for every `1^@"C""` increase in its temperature!

So, for example, your sample of gold is said to have a mass of `"20 g"`. What if we wanted to increase the temperature of this sample by just `1^@"C"?`

Well, since you need `"0.128 J"` for every gram, and since you only want to increase its temperature by `1^@C"`, it follows that you'd need

`20 xx "0.128 J" = "2.56 J"`

But you ended up using `"96 J"`, which means that the difference between this value and the value given to you went into increasing the sample's temperature!

You can say that

`x xx "2.56 J" = "96 J"`

Here `x` represents the increase in temperature produced by adding that extra heat to the sample. This means that you have

`x = 96/2.56 = 37.5`

The temperature of the coin increased by `37.5^@"C"`.

You actually have an equation that you can use for such problems

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

`q` - the amount of heat added / removed
`m` - the mass of the sample
`c` - the specific heat of the substance
`DeltaT` - the change in temperature

Plugging your values into this equation will once again give you

`DeltaT = q/(m * c)`

`DeltaT = (96 color(red)(cancel(color(black)("J"))))/(20 color(red)(cancel(color(black)("g"))) * 0.128color(red)(cancel(color(black)("J")))/(color(red)(cancel(color(black)("g"))) ""^@"C")) = 37.5^@"C"`

I'll leave the answer rounded to two sig figs

`DeltaT = color(green)(38^@"C")`