Cliffs AP Chemistry, 3rd Edition

Together, the products are absorbing 81 J/°C + 54 J/°C = 135 joules for every l°C rise in

temperature. Because the reaction is producing 847.6 kJ of energy (you know this from

your answer to part b), the change in temperature, from the initial conditions of the reac-

tants (presumably at room temperature) to those of the hot products, is found by dividing

the heat of reaction by the energy absorbed per degree Celsius (J divided by J/°C = °C).

Thus

/ C

,

135.

847 600

JC^62810

J

% = #^3 % change in temperature

(d) Restatement: Confirmation that the heat produced from the thermite reaction is suffi-

cient to melt iron.

Given: HfusFe = 270 J/g

Because ∆Hfusrepresents the amount of energy absorbed in melting iron, you must sub-

tract this component from the ∆H you obtained in part (b). And because ∆H reflects the

amount of heat produced when 2 moles of iron are produced, ∆Hfusfor the reaction is

.

1 ,

2

1

55 85

1

moles of Fe (^270) 30 200
mole of Fe
gFe
gFe
##J= J
The 30,200 J of energy was absorbed in melting iron, so you subtract it from the ∆H value
obtained in part (b) to find how much energy was actually released in the reaction.
847,600 J −30,200 J = 817,400 J
Finally, from part (c), you know that the amount of energy released divided by the energy
absorbed per degree Celsius (J divided by J/°C = °C) equals the change in temperature.
Thus you have
C
C
/
,
.
135
817 400
605 10
J
J 3
% = # %
which is above the melting point of iron, 1535°C.

2. You pack six aluminum cans of cola in a cooler filled with ice. Each aluminum can
   (empty) weighs 50.0 grams. When filled, each can contains 355 mL of cola. The density

of the cola is 1.23 g/mL. The specific heat of aluminum is 0.902 J/g ⋅°C, and that of the

cola is 4.00 J/g ⋅°C.

(a) If the cola is initially at 30°C and you wish to cool it to 10°C, how much heat must

be absorbed by the ice?

(b) What is the minimum amount of ice (at 0°C) needed to cool the cola? ∆Hfusfor ice is

6.00 kJ/mole.

Answer

2. Given: 6 aluminum cans (empty) at 50.0 g each
   each can = 355 mL cola
   density of cola = 1.23 g/mL

Part II: Specific Topics