Answer:
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
1 gram
0.90 J
+
1 gram
0.90 J
+
...
+
1 gram
0.90 J
10 times
=
10
×
0.90 J
However, you don't want to increase the temperature of the sample by
1
∘
C
, you want to increase it by
Δ
T
=
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
1
∘
C
10
×
0.90 J
+
1
∘
C
10
×
0.90 J
+
...
+
1
∘
C
10
×
0.90 J
33 times
=
33
×
10
×
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
g
⋅
0.90
J
g
∘
C
⋅
33
∘
C
q
=
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
q
=
m
⋅
c
⋅
Δ
T
, where
q
- the amount of heat added / removed
m
- the mass of the substance
c
- the specific heat of the substance
Δ
T
- the change in temperature, defined as the difference between the final temperature and the initial temperature of the sample
