Question

How many joules of heat are needed to raise the temperature of 10.0 g of aluminum from 22°C to 55°C, if the specific heat of aluminum is 0.90 J/g°C?

Answer 1

`"297 J"`

Explanation:

The key to this problem lies with aluminium's specific heat, which as you know tells you how much heat is needed in order to increase the temperature of `"1 g"` of a given substance by `1^@"C"`.

In your case, aluminium is said to have a specific heat of `0.90 "J"/("g" ""^@"C")`.

So, what does that tell you?

In order to increase the temperature of `"1 g"` of aluminium by `1^@"C"`, you need to provide it with `"0.90 J"` of heat.

But remember, this is how much you need to provide for every gram of aluminium in order to increase its temperature by `1^@"C"`. So if you wanted to increase the temperature of `"10.0 g"` of aluminium by `1^@"C"`, you'd have to provide it with

`underbrace(overbrace("0.90 J")^(color(blue)("1 gram")) + overbrace("0.90 J")^(color(blue)("1 gram")) + " ... " + overbrace("0.90 J")^(color(blue)("1 gram")))_(color(red)("10 times")) = 10 xx "0.90 J"`

However, you don't want to increase the temperature of the sample by `1^@"C"`, you want to increase it by

`DeltaT = 55^@"C" - 22^@"C" = 33^@"C"`

This means that you're going to have to use that much heat for every degree Celsius you want the temperature to change. You can thus say that

`underbrace(overbrace(10 xx "0.90 J")^(color(purple)(1^@"C")) + overbrace(10 xx "0.90 J")^(color(purple)(1^@"C")) + " ... " + overbrace(10 xx "0.90 J")^(color(purple)(1^@"C")))_(color(red)("33 times")) = 33 xx 10 xx "0.90 J"`

Therefore, the total amount of heat needed to increase the temperature of `"10.0 g"` of aluminium by `33^@"C"` will be

`q = 10.0 color(red)(cancel(color(black)("g"))) * 0.90"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * 33color(red)(cancel(color(black)(""^@"C")))`

`q = color(green)("297 J")`

I'll leave the answer rounded to three sig figs, despite the fact that your values only justify two sig figs.

For future reference, this equation will come in handy

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

`q` - the amount of heat added / removed
`m` - the mass of the substance
`c` - the specific heat of the substance
`DeltaT` - the change in temperature, defined as the difference between the final temperature and the initial temperature of the sample