2. The heat of combustion of octane, CgH18, can be measured in a way similar to the method you used to measure the heat of neutralization. The bomb calorimeter is used instead of a coffee cup. The reaction is 2 CHi ()25 O2 (g) 16 CO2 (g 18 H20 (1) When 1.02 g of octane was burned in a bomb calorimeter, the temperature of 1.00 kg of water was raised from 22.0 °C to 35.3 °C. The specific heat of water is 4.184 J/ (g °C) How many moles of octane were burned? What is the heat of combustion H) of octane in kilojoules per mole? 3. In the calorimetry lab how would your results have been affected if you had not used a lid to cover the polystyrene cup after the acid and base solutions had been mixed? What would have been the effect on the measurement of temperature and why? 4. Define the following terms a) specific heat b) heat capacity 5. A coffee cup calorimeter contains 25.0 g water at 23.8 C. A 5.00 g sample of an unknown metal at an initial temperature of 78.3 C was dropped into the calorimeter. The final temperature of mixture was 46.3 °C. Calculate the specific heat of the metal. The specific heat of water is 4.184 J/ (g °C) 6. In a coffee cup, 75.0 g water initially at 23.0 C was mixed with 25.0 g of water initially at 95.0 °C. What will the final temperature be after mixing? Solution 2. Heat absorbed by water: H = m × specfic heat of water × (Final temperature - Initial temperature) = 1000 g × 4.184 J/ g 0 C × (35.3 - 22) 0 C =55647 J = 55.647 kJ This amount of heat is released by combustion of 1.02 g octane. Number of moles of octane = 1.02 g/ 114 g/ mole = 0.009 mole Heat of combustion of octane : 55.647 kJ / 0.009 mole = 6183 kj/ mole 3. If lif was opened then some heat would transfer to environment. So; the water in the calorimeter does not absorb all the heat produced by reaction. Rather it absorbs part of heat. So the temperature rise will be less than the actual temperature rise (if the lid was closed). 4. (a) Specific heat of an object is the heat required to raise the temperature of unit mass of the object by one degree . Heat capacity is the heat required to raise the temperature of a soecific substance by one degree. 5. Heat gained by water : H1 = 25 g × 4.184 J/ g 0 C ×( 46.3-23.8) 0 C = 2353.5 Joules Heat released by metal : H2 = 5 g × s J/g 0 C × ( 46.3 - 78.3) 0 C = -160×s J oules At thermal equilibrium; H1 = -H2 So; 2353.5 = 160×s Or; s = 14.71 Specific heat of the metal is 14.71 J/ g 0 C. .

1. 2. The heat of combustion of octane, CgH18, can be measured in a way similar to the method
you used to measure the heat of neutralization. The bomb calorimeter is used instead of a coffee
cup. The reaction is 2 CHi ()25 O2 (g) 16 CO2 (g 18 H20 (1) When 1.02 g of octane was burned
in a bomb calorimeter, the temperature of 1.00 kg of water was raised from 22.0 °C to 35.3
°C. The specific heat of water is 4.184 J/ (g °C) How many moles of octane were burned?
What is the heat of combustion H) of octane in kilojoules per mole? 3. In the calorimetry lab
how would your results have been affected if you had not used a lid to cover the polystyrene cup
after the acid and base solutions had been mixed? What would have been the effect on the
measurement of temperature and why? 4. Define the following terms a) specific heat b) heat
capacity 5. A coffee cup calorimeter contains 25.0 g water at 23.8 C. A 5.00 g sample of an
unknown metal at an initial temperature of 78.3 C was dropped into the calorimeter. The final
temperature of mixture was 46.3 °C. Calculate the specific heat of the metal. The specific heat
of water is 4.184 J/ (g °C) 6. In a coffee cup, 75.0 g water initially at 23.0 C was mixed with
25.0 g of water initially at 95.0 °C. What will the final temperature be after mixing?
Solution
2. Heat absorbed by water: H = m × specfic heat of water × (Final temperature - Initial
temperature) = 1000 g × 4.184 J/ g 0
C × (35.3 - 22) 0
C =55647 J = 55.647 kJ
This amount of heat is released by combustion of 1.02 g octane. Number of moles of octane =
1.02 g/ 114 g/ mole = 0.009 mole
Heat of combustion of octane : 55.647 kJ / 0.009 mole = 6183 kj/ mole
3. If lif was opened then some heat would transfer to environment. So; the water in the
calorimeter does not absorb all the heat produced by reaction. Rather it absorbs part of heat. So
the temperature rise will be less than the actual temperature rise (if the lid was closed).
4.
(a) Specific heat of an object is the heat required to raise the temperature of unit mass of the
object by one degree .
Heat capacity is the heat required to raise the temperature of a soecific substance by one degree.

2. 5. Heat gained by water : H1 = 25 g × 4.184 J/ g 0
C ×( 46.3-23.8) 0
C = 2353.5 Joules
Heat released by metal : H2 = 5 g × s J/g 0
C × ( 46.3 - 78.3) 0
C = -160×s J oules
At thermal equilibrium; H1 = -H2
So; 2353.5 = 160×s
Or; s = 14.71
Specific heat of the metal is 14.71 J/ g 0
C.