Calculate the enthalpy change, delta H_1 for the process in which 37.7 g of wate
ID: 978823 • Letter: C
Question
Calculate the enthalpy change, delta H_1 for the process in which 37.7 g of water is converted from liquid at 19.9 degree C to vapor at 25.0 degree C. For water, delta H =44.0 kJ/mol at 25.0 degree C and C =4.18 J/(g. degree C) for H_2O(l). Express your answer to three significant figures and include the appropriate units. How many grams of ice at -10.3 degree C can be completely connected to liquid at 21.0 degree C if the available heat for this process is 5.53 times 10^3 kJ? For ice, use a specific heat of 2.01 J/(g. degree C) and delta H =6.01 kJ/mol. Express your answer to three significant figures and include the appropriate units.Explanation / Answer
Calculate the enthalpy change for the process in which 37.7 g of water
Temperature change = 25.0-19.9= 5.1
The total enthalpy is including two processes:
Q1= heat the water from 19.9ºC to 25.0ºC,
Q1 = mass x specific heat x deltaT
Q1= 37.7 g x 4.18J/g-ºC x 5.1 ºC
= 804.46 J
=0.804 KJ
Moles of water = 37.7 g/ 18.02 g/ mol= 2.09 mol
Q2=heat the water to steam at 25ºC
Q2= moles water x heat of vaporization
heat = 2.09mol x 44.0 kJ/mole
= 9205 kJ
total heat;Q
= 9205 kJ + 0.804 KJ
= 9205 kJ .804 kJ
b) Q1 = mass x specific heat x deltaT
Q 2 = moles ice x molar heat of fusion
Q 3 = mass x specific heat water x deltaT water
total heat ;Q = Q 1 + Q 2 + heat Q 3
assume that x = mass ice
As deltaT in kelvin = deltaT in C, we can use the deltaT in C as the deltaT in K
5530 kJ = (x)(2.108 kJ/kg-K)(10.3K) + (x/18.02g/mole)(333.55 kJ/kg) + (x)(4.18J/g-ºC)(21.0K)
5530 = 21.7 x + 18.51x + 87.78x
5530= 127.99 x
x= 43.21 kg
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