Calorimetry - worked problems PDF

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Calorimetry Problems 1. A water calorimeter at 20.1oC contains 643.2 g of water. Addition of 2.744 g of KOH to the calorimeter produces a temperature rise to 23.4oC. Let us compute the molar heat of solution of KOH from these data. The molar heat capacity of water is 75.291 J/mol K.

2: Suppose a piece of iron with a mass of 21.5 g at a temp of 100.0 °C is dropped into an insulated container of water. The mass of the water is 132.0 g and its temperature before adding the iron is 20.0 °C. What will be the final temp of the system? The specific heat of iron is 0.449 kJ / kg K.

3. 100 g of nickel at 150 oC is placed in 1 L of water at 25 oC. The final temperature of the water is 26.3 oC. Given that the heat capacity of water is 4.18 J g-1 K-1, what is the specific heat of nickel?

4. A certain material’s temperature increases by 1 K for every 1560 J that it gains. A 0.1964 g sample of quinone (molar mass = 108.1 g/mole) was burnt, and the surrounding material’s temperature increased from 20.3 oC to 23.5 oC. Find the molar heat of combustion for quinone.

5. A 50 g sample of 0.2 M NaBr aqueous solution at 23.65 oC is added to a styrofoam calorimeter containing 50 g of 0.2 M AgNO3 aqueous solution at 23.65 oC. If the heat capacity of the calorimeter is 65 J K-1 and the specific heat of the solution is 4.20 J g-1 K-1 and the final temperature of the solution in the calorimeter is 25.40 oC, calculate the heat released in the reaction.

Calorimetry Solutions: 1. Answer: 54.2 kJ/mol The temperature rise, or difference between the initial and final temperatures, is +3.3oC or +3.3 K. Ignoring the solute, the calorimeter contains 634.2 g of water which is 634.2 g/(18.015 g/mol) = 35.204 mol water. The molar heat capacity of water is 75.291 J/mol K so the total heat of dissolution is 2.651 kJ. The heat was produced by dissolution of (2.744 g/56.109 g KOH/mol KOH) = 0.0489 mol of KOH, so the molar heat of solution of KOH is (2.651 kJ)/(0.0489 mol) = 54.2 kJ/mol.

2. Answer: 21.4 °C. Since qlost, metal = qgained, water we write (mass) (Δt) (Cp, metal) = (mass) (Δt) (Cp, water) Substituting: (21.5) (100 - x) (0.449) = (132.0) (x - 20) (4.184) Some explanation: a) 100 - x is the Δt for the metal; it starts at 100.0 °C and drops to some unknown, final value. b) x - 20 is the Δt for the water; it starts at 20.0 °C and rises to some unknown, final value. c) Since both metal and water wind up at the same ending value, we need to use only one unknown for the two Δt expressions.

3. Answer: 0.44 J g-1 K-1

4. Answer: -2.7 x 103 kJ / mol Qmaterial = 1560 J/K (23.5-20.3) K = 49 92 J = 5.0 kJ Note that 1560 J/K is equivalent to mc from Q = mcDT. H = -Q = -5.0 kJ n = 0.1964g/[108.1 g/mole] = 0.00182 mol H/n = -5.0 kJ/0.00182 mol = -2.7 X 103 kJ/mol of quinine

5. Answer: 8.49 x 102 J Heat released in the reaction = sum of heat absorbed by the calorimeter and the solution Heat absorbed by the solution is given by the product of the specific heat of the solution, the mass of the solution and the change in temperature of the solution. Heat absorbed by the calorimeter is given by the product of the heat capacity of the calorimeter and the change in temperature of the calorimeter....