The beaker of acetic acid will cool more quickly.
The specific heat capacity of acetic acid is about half that of water.
Thus, it takes twice as much heat gain (or loss) in acetic acid to cause a given change in temperature.
If everything else is constant and heat is being lost at the same rate, the temperature of the acetic acid should drop twice as fast as that of water.
LET'S PUT IN WHAT WE KNOW!!!
Q
=
725 J
m
=
55.0 g
c
=
0.900 J/(°C⋅g)
Δ
T
=
final temperature - initial temperature
Δ
T
=
(
x
−
27.5
)
°C
We solve for
Δ
T
.
725 J
=
55.0 g
⋅
0.900 J/(°C⋅g)
(
x
−
27.5
)
°C
NOW IT'S JUST BASIC ALGREBRA
725
=
49.5
x
−
1361
2086
=
49.5
x
42.1
=
x
The final temperature is 42.1 °C.
Answer:
5.72 g ( 1 mol / 180.16 g ) ( 6.022 x 10^23 molecules / mole ) = 1.90x10^23 molecules
X is always the independent variable
