specific heat of aluminum = 0.902 J/g ⋅°C
specific heat of cola = 4.00 J/g ⋅°C
ti= 30 °C
tf= 10°C
(a) Restatement: Amount of heat absorbed by ice.
quantity of heat released by six pack = heat released by cans + cola
qsix-pack= qcans+ qcola
q= mass ×specific heat ×∆t
C
C
g
g
:
:
..
.
..
.
.
.,
q
q
q
1
6
1
50 0 0 902
1
10 30
5412 00
1
6
1
355
1
(^123400)
1
10 30
210 10
5412 2 10 10
2 15 10 215
cans
can
gAl J CC
J
cans
can
mL cola
mL cola
g cola J CC
J
JJ
Jor kJ
5
5
5
cans
cola
six pack
=
=-
=
=-
=- + -
=- -
%
%
%
%
%
%
`
`
_
j
j
i
Therefore, the amount of heat absorbed by the ice is 215 kJ.
(b) Restatement: Minimum amount of ice required to accomplish the necessary cooling.
For the systemto reach 10°C, the heat absorbed includesthe warming of the water.
Let x = mass of ice
././. CC
./
x g
x
(^215) 18 02^600 4 184 10
0 375
(^215573)
kJ kJ molemole J/g
kJ g
kJ g ice
=+:#
d nH ` %%j
Thermochemistry