1. a) 50ml of water at 46.9°C were mixed with 50ml of water at 25.1°C in a calorimeter also at 25.1°C. The final temperature was 30.1°C. Assuming that neither the density of water nor its specific heat capacity change with temperature, calculate the total heat capacity of the calorimeter. (density of water = 1.00g/ml, specific heat capacity = 4.18 j/gk

Question

1. a) 50ml of water at 46.9°C were mixed with 50ml of water at 25.1°C in a calorimeter also at 25.1°C. The final temperature was 30.1°C. Assuming that neither the density of water nor its specific heat capacity change with temperature, calculate the total heat capacity of the calorimeter. (density of water = 1.00g/ml, specific heat capacity = 4.18 j/gk

b) when 5.00g of NaOH (s) are added to 100g of water (using the same calorimeter as in part a), the temperature rises from 25.0°C to 37.5°C. calculate the molar heat of solution, i.e. delta H for the process. assume that the specific heat of water is 4.18 j/gk and that of the NaOH (aq) solution is the same.

Answer 1

a.

[mass warm H2O x specific heat H2O x (Tfinal-Tinitial)] + [mass cool H2O x specific heat H2O x (Tfinal-Tinitial)] ]+ [Ccal x (Tfinal-Tinitial)] = 0
Solve for Ccal.

b.
[mass H2O x specific heat H2O x (Tfinal-Tinitial)] + [Ccal x (Tfinal-Tintial)] = q = dHsoln in joules.
Convert 5.00 g NaOH to mols. mols = grams/molar mass.

Then dHsoln = q/mol NaOH = ?J/mol. These are most often quoted in kJ/mol and you can make that conversion; divide J/mol by 1000 for kJ/mol

Answer 2

a) To calculate the total heat capacity of the calorimeter, we need to use the principle of energy conservation. The heat lost by the warmer water will be equal to the heat gained by the cooler water and the calorimeter.

1. Calculate the heat lost by the warmer water:
Q1 = mass1 * specific heat capacity1 * temperature change1
Q1 = 50g * 4.18 J/g°C * (46.9°C - 30.1°C)

2. Calculate the heat gained by the cooler water:
Q2 = mass2 * specific heat capacity2 * temperature change2
Q2 = 50g * 4.18 J/g°C * (30.1°C - 25.1°C)

3. Since the heat lost and gained should be equal:
Q1 = Q2 + Q_calorimeter

Assuming the heat capacity of the calorimeter mass (C_calorimeter) is the same as the water:
Q_calorimeter = mass3 * specific heat capacity3 * temperature change3
Q_calorimeter = 100g * 4.18 J/g°C * (30.1°C - 25.1°C)

Q1 = Q2 + Q_calorimeter
(50g * 4.18 J/g°C * (46.9°C - 30.1°C)) = (50g * 4.18 J/g°C * (30.1°C - 25.1°C)) + (100g * 4.18 J/g°C * (30.1°C - 25.1°C))

Solve the equation to find the value of C_calorimeter, which represents the total heat capacity of the calorimeter.

b) To calculate the molar heat of solution, we need to use the equation:

ΔH = Q / moles

1. Calculate the heat absorbed by the water:
Q_water = mass_water * specific heat capacity_water * temperature change_water
Q_water = 100g * 4.18 J/g°C * (37.5°C - 25.0°C)

2. Calculate the heat absorbed by the NaOH (solute):
Q_NaOH = mass_NaOH * molar heat capacity_NaOH * temperature change_NaOH
First, calculate the moles of NaOH:
Moles_NaOH = mass_NaOH / molar mass_NaOH
Next, calculate the heat absorbed:
Q_NaOH = Moles_NaOH * ΔH

3. Since the heat absorbed by the water and the NaOH will be equal:
Q_water = Q_NaOH

Solve the equation for ΔH, which represents the molar heat of solution.

Note: The molar heat capacity of NaOH, specific heat capacity of water, and molar mass of NaOH are required for the calculation.