1.How many milliliters of water at 25.0°C with a density of 0.997 g/mL must be m

Question

1.How many milliliters of water at 25.0°C with a density of 0.997 g/mL must be mixed with 181 mL of coffee at 97.2°C so that the resulting combination will have a temperature of 50.4°C? Assume that coffee and water have the same density and the same specific heat (4.18 J/g·°C) across the temperature range.

2.How many moles of butane, C4H10(l), must be burned to produce 182 kJ of heat under standard state conditions? (The heat of combustion for butane is -2855.7 kj/mol)

3.When a chemist mixed 3.32 g of LiOH and 255 mL of 0.65 M HCl in a constant-pressure calorimeter, the final temperature of the mixture was 24.5°C. Both the HCl and LiOH had the same initial temperature, 21.5°C. The equation for this neutralization reaction is:

Given that the density of each solution is 1.00 g/mL and the specific heat of the final solution is 4.1801 J/g·K, calculate the enthalpy change for this reaction in kJ/mol LiOH. Assume no heat is lost to the surroundings.

4.A foundry worker places a 5.77 kg sheet of nickel at a temperature of 16°C on top of a 16.7 kg sheet of lead at 65°C. Assuming no heat is lost to the surroundings, calculate the final temperature of the two sheets of metal.

Explanation / Answer

1.

heat lost by coffee = heat gained by water

heat = mass x specific heat x change in temperature

mass coffee = 181 ml x 1 g/ml = 181 g

181 x 4.18 x (97.2 - 50.4) = (m/0.997) x 4.18 x (50.4 - 25)

Volume of water to be added = 332.5 ml

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2. 2C4H10 + 13O2 --> 8CO2 + 10H2O                     dHcomb = -2855.7 kJ/mol

To produce -182 kJ heat,

moles C4H10 needed = 2 x -182/-2855.7 = 0.13 moles

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3. Total mass (m) = 3.32 + 255 = 258.32 g

change in temperature dT = 24.5 - 21.5 = 3.0 oC

specific heat Cp = 4.1801 J/g.K

enthalpy change = -mCpdT/moles LiOH

                            = -258.32 x 4.1801 x 3 x 23.95/3.32 x 1000

                            = -23.37 kJ/mol

4. Let the final temperature be Tf

heat lost by lead = heat gained by nickel

16.7 x 1000 x 4.184 x (65 - Tf) = 5.77 x 1000 x 4.184 x (Tf - 16)

1085500 - 16700Tf = 5770Tf - 92320

Tf = 1177820/22470 = 52.42 oC

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