Question

Question #5dc43

Answer 1

`"0.015 kg Al"`

Explanation:

The key to this problem is aluminium's specific heat, which tells you how much heat is needed to increase the temperature of `"1 g"` of aluminium by `1^@"C"`.

Aluminium's specific heat is said to be equal to `"0.897 J g"^(-1)""^@"C"^(-1)`. This tells you that in order to increase the temperature of `"1 g"` of aluminium by `1^@"C"`, you need to supply it with `"0.897 J"` of heat.

Now, let's assume that your sample has a mass of `"1 g"`. In order to increase its temperature by

`DeltaT = 187^@"C" - 65^@"C" = 122^@"C"`

you would need to provide it with `122` times more heat than what would be needed to increase its temperature by `1^@"C"`.

More specifically, you would need to provide it with

`122color(red)(cancel(color(black)(""^@"C"))) * overbrace("0.897 J"/(1color(red)(cancel(color(black)(""^@"C")))))^(color(purple)("heat needed for"color(white)(a)1^@"C"color(white)(a) "increase for 1 g of Al")) = "109.434 J"`

However, you end up adding `"1650 J"` of heat to this sample, which of course means that its mass is not equal to `"1 g"`.

The ratio between the heat added to the sample to get its temperature to increase by `122^@"C"` and the heat that would be needed to increase the temperature of `"1 g"` of aluminium by `122^@"C"` will give you the mass of the sample, expressed in grams

`1650color(red)(cancel(color(black)("J"))) * overbrace("1 g"/(109.434color(red)(cancel(color(black)("J")))))^(color(purple)("heat needed for"color(white)(a) 122^@"C" color(white)(a)"increase for 1 g of Al")) = "15.08 g"`

Expressed in kilograms and rounded to two sig figs, the answer will be

`color(purple)(|bar(ul(color(white)(a/a)color(black)("1 kg" 1= 10^3"g")color(white)(a/a)|)))`

`"mass of Al" = color(green)(|bar(ul(color(white)(a/a)"0.015 kg"color(white)(a/a)|)))`

ALTERNATIVE SOLUTION

You can get the same result by using the formula

`color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))" "`, where

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

Rearrange to solve for `m`

`q = m * c * DeltaT implies m = q/(c * DeltaT)`

Plug in your values to get

`m = (1650 color(red)(cancel(color(black)("J"))))/(0.897color(red)(cancel(color(black)("J"))) "g" color(red)(cancel(color(black)(""^@"C"^(-1)))) * (187 - 65)color(red)(cancel(color(black)(""^@"C"))))`

`m = "15.08 g"`

Once again, you'll have

`"mass of Al" = color(green)(|bar(ul(color(white)(a/a)"0.015 kg"color(white)(a/a)|)))`