Calorimetry Problem: Show your work neatly and methodically. Include the sign as

Question

Calorimetry Problem: Show your work neatly and methodically. Include the sign associated with H.

1. When a 6.55 gram sample of solid sodium hydroxide dissolves in 115.00 grams of water in a coffee-cup calorimeter, the temperature rises from 21.6°C to 38.7°C. Calculate H, in kJ/mole NaOH, for the solution process. NaOH(s) Na1+(aq) + OH1- (aq) The specific heat of the solution is 4.18 J/g °C. 3.

2, A 2.600 gram sample of phenol, C6H5OH, was burned in a bomb calorimeter whose total heat capacity was 11.66 kJ/°C. The temperature of the calorimeter increases from 21.36°C to 30.57°C. What is the heat of combustion per mole of phenol?

3. A large cup contains 150.0 grams of water at 25.1°C. A 121.0 gram block of copper metal is heated to 100.4°C. The 100.4°C copper block is added to the 25.1°C water in the cup. All of the heat lost by the copper block was absorbed by the water in the cup. After a time, the contents of the cup reaches a constant temperature. What is the final temperature of the contents of the cup? The specific heat of water is 4.18 J/g °C. The specific heat of copper metal is 0.385 J/g °C.

4. When a 3.88 gram sample of solid ammonium nitrate dissolves in 60.0 grams of water in a coffee-cup calorimeter, the temperature drops from 23.0°C to 18.4°C. Calculate H, in kJ/mole NH4NO3, for the solution process: Specific heat of the solution = 4.18 J/g-K. NH4NO3(s) NH4 1+(aq) + NO3 1- (aq) Include the sign of H. Is this process endothermic or exothermic?

Explanation / Answer

1) mass of water = m =115 g

dT = 38.7 -21.6 = 17.1 oC

Cp = 4.18 J / g oC

Q = m Cp dT = 115 x 4.18 x 17.1 = 8220 J = 8.220 kJ

moles of NaOH = 6.55 / 40 = 0.164

H = - Q / n = 8.220 / 0.164

= - 50 .12 kJ / mol

2)

Q = Cp x dT = 11.66 x (30.57 -21.36) = 107.39 kJ

moles of phenol = 2.6 / 94 = 0.0277

heat of combustion = - 107.39 / 0.0277

= -3882.5 kJ / mol

3)

heat loss copper block = heat gained by water

(m Cp dT ) metal = (m Cp dT ) water

121 x 0.385 x (100.4 -Tf) = 150 x 4.18 x (Tf -25.1)

Tf = 30.3 oC

final temperature = 30.3 oC

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