1) In a coffee cup calorimeter, 50.0ml of 1.00M NaOH and 50.0ml of 1.00M HCl are

Question

1) In a coffee cup calorimeter, 50.0ml of 1.00M NaOH and 50.0ml of 1.00M HCl are mixed. Both solutions were originally at 24 degrees celsius. After the reaction, the final temperature is 31.3 degrees celsius. Given that density of NaCl solution is 1.038g/ml and the specific heat of NaCl solution is 3.87J/g degrees celsius, calculate the ?H/mole for the reaction of HCl with NaOH. Assume that no heat is lost to the surroundings.

2) The molar mass of silver is 108 grams per mole. The specific heat of silver is 0.28J/g degrees celsius. What is the molar heat capacity of silver?

3) A 0.0100 mole sample of butane was burned in a  bomb calorimeter. The heat from the reaction raised the temperature of 1.00kg of water by 6.90 degrees celsius. How much heat was released in this reaction per mole of butane combusted? The specific heat of water is 4.184J/(g degrees celsius).

4) The combustion of 0.1584 grams of benzoic acid increases the temperature of a calorimeter by 2.54 degrees celsius. Calculate the heat capacity of this calorimeter. The energy released by the combustion of benzoic acid is 26.42kj/gram.

5) Suppose the heat of neutralization had been determined using a glass beaker instead of a polystyrene coffee cup. How would the measured value have been affected?

6) The heat of combustion of octane, C8H18, 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: 2C8H18(l)+25O2(g)-> 16CO2(g)+18H2O(1) When 1.02g of octane was burned in a bomb calorimeter, the temperature of 1.00kg of water was raised from 22.0 degrees celsius to 35.3 degrees celsius. The specific heat of water is 4.184J?(g degrees celsius). How many moles of octane were burned? What is the heat combustion (H) of octane in kilojoules per mole?

7) 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?

8) A coffee cup calorimeter contains 25.0g water at 23.8 degrees celsius. A 5.00g sample of an unknown metal at an initial temperature of 78.3 degrees celsius was dropped into the calorimeter. The final temperature of mixture was 46.3 degrees celsius. Calculate the specific heat of the metal. The specific heat of water is 4.184J/(g degrees celsius).

9) In a coffee cup, 75.0g water initially at 23.0 degrees celsius was mixed with 25.0g of water initially at 95.0 degrees celsius. What will the final temperature be after mixing?

Explanation / Answer

9) In a coffee cup, 75.0g water initially at 23.0 degrees celsius was mixed with 25.0g of water initially at 95.0 degrees celsius. What will the final temperature be after mixing?

Heat gained = heat lost

mass * change in temp = mass * change in temp
let final temp be X

75 *(X-23) = 25*(95-X)

75X - 1725 = 2375 -25X

100X = 4100

X = 41

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