How much heat is required in kJ to convert 26.9 g of water at 97.5°C to steam at
ID: 573993 • Letter: H
Question
How much heat is required in kJ to convert 26.9 g of water at 97.5°C to steam at 105.5°C? The boiling point of water is 100.0°C, Cm for liquid water = 75.4 J/(mol•°C), Hvap = 40.67 kJ/mol, and Cm for steam = 33.6 J/(mol•°C).
An unknown liquid has a boiling point of 617.35 ºC, and the enthalpy change for the conversion of this liquid to a gas is Hvap = 56.2 kJ/mol. What is the entropy change for vaporization, Svap, in J/(K•mol)?
The vapor pressure of ethanol at 306.7 K is 100. mmHg, and the heat of vaporization of ethanol is 38.6 kJ/mol. What is the vapor pressure of ethanol in mmHg at 330.1 K?
A newly discovered element with a molecular weight of 328.8 g/mol crystallizes in a simple cubic arrangement, with the edge of a unit cell having a length d = 387.9 pm. What is the density of this element in g/cm3?
Explanation / Answer
1)
97.5 -------- 100 --------- 105.5 oC
mass of water = 26.9 g
moles of water = 26.9 / 18.02 = 1.493
Q1 = m Cp dT
= 1.493 x 75.4 x (100 - 97.5)
= 281.39 J
Q2 = 1.493 x 40.67 = 60720.3 J
Q3 = 1.493 x 33.6 x (105.5 - 100)
= 275.9 J
total heat required = Q1 + Q2 + Q3
= 61277.6 J
total heat required = 61.3 kJ
2)
temperature = 617.35 oC = 890.5 K
Svap = Hvap / T
= 56.2 x 10^3 / 890.5
Svap = 63.11 J / K.mol
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