Question

Question #6ad9d

Answer 1

`"2250 J"`

Explanation:

The key to this problem is the specific heat of aluminium

`c_"Al" = color(purple)("0.899 J")/(color(green)("g")color(blue)(""^@"C"))`

The specific heat of a substance tells you how much heat is required to reaise the temperature of `"1 g"` of said substance by `1^@"C"`.

In your case, you know that you need `color(purple)("0.899 J"` of heat in order to increasde the temperature of `color(green)("1 g"` of aluminium by `color(blue)(1^@"C")`.

You can use the specific heat of aluminium to calculate how much heat would be required to raise the temperature of `"500.0 g"` of aluminium by `1^@"C"`

`500.0 color(red)(cancel(color(black)("g"))) * color(purple)("0.899 J")/(color(green)(1)color(red)(cancel(color(green)("g"))) color(blue)(1^@"C")) = "449.5 J"color(blue)(""^@"C"^(-1))`

This tells you that you need to provide `"449.5 J"` in order to raise the heat of `"500.0 g"` of aluminium by `1^@"C"`. The problem tells you that the temperature of the sample must increase by

`DeltaT = 20.0^@"C" - 15.0^@"C" = 5.0^@"C"`

In order to get thatto happen, you must provide it with

`5.0 color(red)(cancel(color(black)(""^@"C"))) * "449.5 J"/(color(blue)(1)color(red)(cancel(color(blue)(""^@"C")))) = ul("2250 J")`

The answer is rounded to three sig figs.

`color(white)(a)`
ALTERNATIVELY

you can also use the following equation

`color(blue)(ul(color(black)(q = m * c * DeltaT)))`

Here

• `q` is the amount of heat gained
• `m` is the mass of the sample
• `c` is the specific heat of the substance
• `DeltaT` is the change in temperature, defined as the difference between the final temperature and the initial temperature

In your case, you will once again end up with

`q = 500.0 color(red)(cancel(color(black)("g"))) * "0.899 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * (20.0 - 15.0) color(red)(cancel(color(black)(""^@"C")))`

`q = color(darkgreen)(ul(color(black)("2250 J")))`