How much heat would be relaesed by the condensation of 5.40g of steam at 100 (degrees)C and the subsequent cooling of the water to 25(degrees)C? [(delta)Hvap=40.7 KJ/mol at 100(degrees)C; Cp for H2O(l) is 4.18Jg-1 (degrees)c-1]
A) 12.2 kj
B) 18.3 kj
C) 12.8 kj
D) 13.9 kj
E) 23.7 kj
q=sm(final-initial temperature).
s= 4.18
m=5.40g
q1 to condense steam at 100 C.
q1 = mass x heat vap
q2 = heat released cooling from 100 C to 25 C.
q2 = mass x specific heat x (Tfinal-Tinitial)
Total q = q1 + q2.
To find the amount of heat released, we need to calculate the heat released during the condensation of steam and the subsequent cooling of the water.
First, we need to calculate the heat released during the condensation of steam, which can be determined using the heat of vaporization (ΔHvap) formula:
Q1 = mass of steam × ΔHvap
Given:
mass of steam = 5.40 g
ΔHvap = 40.7 kJ/mol
To convert grams to moles, we need to know the molar mass of water (H2O):
Molar mass of H2O = 18.015 g/mol
Number of moles of steam = mass of steam / molar mass of H2O
Number of moles of steam = 5.40 g / 18.015 g/mol
Now, we can calculate the heat released during the condensation of steam:
Q1 = (5.40 g / 18.015 g/mol) × 40.7 kJ/mol
Next, we need to calculate the heat released during the subsequent cooling of the water. We can use the equation:
Q2 = mass of water × specific heat capacity × temperature change
Given:
mass of water = 5.40 g
specific heat capacity (Cp) = 4.18 J/g°C
temperature change = 100°C - 25°C
Finally, we can calculate the heat released during the subsequent cooling of water:
Q2 = 5.40 g × 4.18 J/g°C × (100°C - 25°C)
To find the total heat released, we add Q1 and Q2:
Total heat released = Q1 + Q2
Now, let's calculate the values:
Q1 = (5.40 g / 18.015 g/mol) × 40.7 kJ/mol
Q1 ≈ 12.14 kJ
Q2 = 5.40 g × 4.18 J/g°C × (100°C - 25°C)
Q2 ≈ 1089.3 J ≈ 1.09 kJ
Total heat released = Q1 + Q2
Total heat released ≈ 12.14 kJ + 1.09 kJ
Total heat released ≈ 13.23 kJ
Rounding off to one decimal place, we find that the amount of heat released is approximately 13.2 kJ. Therefore, the closest option is D) 13.9 kJ.