Hints for Assignment 7


1. Hints. (a) Total Calories = Calories from fat + Calories from protein + Calories from carbohydrate

= mass of fat (9 cal / g) + mass of protein (4 cal / g) + mass carbohydrate (4 cal/g)

(b) 1 cal = 4.184 J; Therefore 1000 cal = 1 kcal = 1 Cal = 4184 J = 4.184 kJ

(c) % energy from fat = 100 x Calories from fat / total Calories
 
 

2. Hints. See worked example on p. 248, Kotz and Treichel 4th editon. Conversion factors are shown above for part (b).
 
 

3. Hints. See Example 6.3 in Kotz and Treichel 4th edition
 
 

4. Worked example. When 1.000 kg water at a temperature of 23.1 C is mixed with 2.500 kg water at an unknown temperature, the final temperature of the resulting mixture is 29.7 C. What was the initial temperature of the second sample of water?

Solution. heat gained by 1.000 kg water = - heat lost by 2.500 kg water
Let T2 be the initial temperature for the 2.500 kg water
heat gained by 1.000 kg water = (1000 g) (4.184 J / g-K) (29.7 - 23.1) K = 2.76 x 104 J (only 2 s.f. are justified - why?)

heat lost by 2.500 kg water = (2500 g) (4.184 J / g-K) (29.7 - T2 ) K

Therefore 2.76 x 104 J = - 10460 (29.7 - T2)
- 2.64 = 29.7 - T2
T2 = 32.3 C
Note that the temperature change is Tfinal- Tinitial for both samples. The final temperatures are the same, the initial temperatures are different. Also remember that the temperature difference in K is the same as the temperature difference in degrees C.
5. Worked example. A piece of rhodium weighing 187 g was heated to 99.7 C and dropped into 142 g of water at a temperature of 16.1 C. What will be the final temperature of the system once the metal and water have come to thermal equilibrium? The specific heat of rhodium is 0.243 J/g-K.
The solution to this problem is set up the same way as above:
heat lost by the rhodium = - heat gained by the water.
If we let T be the final temperature, and remember that the heat capacity of water is 4.184 J/g-K, then

(187 g Rh) (T - 99.7) (0.243 J/g-K) = - (142 g water) (T - 16.1) (4.184 J/g-K)

Multiply through, cancel units, solve for T:

45.44 T - 4530 = -594.1 T + 9565

T = 22.0 C
 
 

6. Hints. Note that the chemical substance (element or compound) is a solid its melting point, and that it is heated beyond its boiling point. Calculate the following heats separately and add them to get the final answer:
  • heat needed to melt the solid
  • heat needed to raise the temperature of the liquid from the melting point to the boiling point
  • heat needed to vaporize the liquid
  • heat needed to raise the temperature of the vapor (gas) from the boiling point to the final temperature
Remember you don't have a mole of the element or compound and watch your units! See Example 6.4 in Kotz and Treichel, 4th edition, for a worked example. (Note: in Example 6.4, heat has to be added to raise the temperature of the solid up to the melting point. This term does not appear in Question 6 because the initial temperature of the solid is the temperature at the melting point).
7. Hints. Calculate the heat lost by the metal as it cools from its starting point to 0 C. Then remember it takes 333 J to melt each gram of ice. Calculate the mass of ice melted.
 
 

8. Worked example. The enthalpy change for the formation of solid sodium bromate from the elements is -334.1 kJ.

(a) Is this reaction endothermic or exothermic? Answer: exothermic (negative sign on delta H means that energy is released)

(b) What is the enthalpy change for the following reaction?

4 Na (s) + 2 Br2 (l) + 6 O2 (g) ---> 4 NaBrO3 (s)
Answer: We now form four moles of NaBrO3 so four times as much energy is released, or -1336.4 kJ

(c) What is the enthalpy change for the following reaction?

2 NaBrO3 (s) ---> 2Na (s) + Br2 (l) + 3 O2 (g)
Answer: Now we are decomposing 2 moles of sodium bromate back to the elements. Formation of sodium bromate was exothermic; decomposition is therefore endothermic. thus the enthalpy change is 2 x (334.1 kJ) = 668.2 kJ.
9. Hint. See the figure on p. 253 of Kotz and Treichel, 4th edition. Make a rough plot of how the temperature of your compound changes as heat is added to it. Remember:
  • while the solid is melting, the temperature stays constant at the melting point
  • once the solid is melted, additional heat raises the temperature of the liquid
  • when the liquid begins to boil, the temperature stays constant at the boiling point
  • when all the liquid has vaporized, additional heat raises the temperature of the vapor

 
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Last revised: August 20 2000 by M. F. Richardson
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