How much heat in kilojoules is evolved in converting 2.00 mol of steam at 145 ∘C to ice at -35.0 ∘C? The heat capacity of steam is 1.84 J/g∘C and that of ice is 2.09 J/g∘C.
To find out how much heat is evolved in this process, we need to calculate the heat involved in each step separately and then sum them up.
Step 1: Heat involved in cooling the steam from 145°C to 100°C and condensing it into water:
To calculate this, we need to know the heat capacity of steam (1.84 J/g°C), the molar mass of water (18.02 g/mol), and the heat of vaporization of water (40.7 kJ/mol).
First, let's calculate the heat involved in cooling the steam from 145°C to 100°C:
Heat_1 = (mass of steam) * (specific heat capacity of steam) * (change in temperature)
Given:
- Mass of steam = 2.00 mol * (molar mass of water) = 2.00 mol * 18.02 g/mol = 36.04 g
- Specific heat capacity of steam = 1.84 J/g°C
- Change in temperature = 100°C - 145°C = -45°C
Heat_1 = 36.04 g * 1.84 J/g°C * (-45°C) = -2985.96 J
Next, let's calculate the heat involved in condensing the steam into water:
Heat_2 = (moles of steam) * (heat of vaporization of water)
Given:
- Moles of steam = 2.00 mol
- Heat of vaporization of water = 40.7 kJ/mol = 40,700 J/mol
Heat_2 = 2.00 mol * 40,700 J/mol = 81,400 J
Total heat involved in Step 1 = Heat_1 + Heat_2 = -2985.96 J + 81,400 J = 78,414.04 J
Step 2: Heat involved in cooling the water from 100°C to 0°C:
To calculate this, we need to know the heat capacity of water (4.18 J/g°C).
Heat_3 = (mass of water) * (specific heat capacity of water) * (change in temperature)
Given:
- Mass of water = 36.04 g (since 1 mol of water is formed from 1 mol of steam)
- Specific heat capacity of water = 4.18 J/g°C
- Change in temperature = 0°C - 100°C = -100°C
Heat_3 = 36.04 g * 4.18 J/g°C * (-100°C) = -15,052.32 J
Step 3: Heat involved in freezing the water into ice at 0°C:
To calculate this, we need to know the heat of fusion of water (6.02 kJ/mol) and the heat capacity of ice (2.09 J/g°C).
Heat_4 = (moles of water) * (heat of fusion of water)
Given:
- Moles of water = 2.00 mol
- Heat of fusion of water = 6.02 kJ/mol = 6,020 J/mol
Heat_4 = 2.00 mol * 6,020 J/mol = 12,040 J
Step 5: Heat involved in cooling the ice from 0°C to -35°C:
To calculate this, we need to know the heat capacity of ice (2.09 J/g°C).
Heat_5 = (mass of ice) * (specific heat capacity of ice) * (change in temperature)
Given:
- Mass of ice = 36.04 g (since 1 mol of water forms 1 mol of ice)
- Specific heat capacity of ice = 2.09 J/g°C
- Change in temperature = -35°C - 0°C = -35°C
Heat_5 = 36.04 g * 2.09 J/g°C * (-35°C) = -2,436.46 J
Total heat involved in Step 5 = Heat_3 + Heat_4 + Heat_5 = -15,052.32 J + 12,040 J - 2,436.46 J = -5,448.78 J
Finally, let's convert the total heat to kilojoules:
Total heat evolved = (Heat in Step 1 + Heat in Step 5) / 1000
Total heat evolved = (78,414.04 J + -5,448.78 J) / 1000 = 72.96 kJ
Therefore, the heat evolved in converting 2.00 mol of steam at 145°C to ice at -35.0°C is approximately 72.96 kilojoules.