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