Calculate the enthalpy change, deltaH, for the process in which 37.7 g of watar

Question

Calculate the enthalpy change, deltaH, for the process in which 37.7 g of watar is coneverted from liquid at 19.9 degC to vapor at 25.0 degC. For water, deltaH_vap = 44.0 kJ/mol al 25.0 degC and C_s = 4.18 J/(g. degC) for H_2O(1). Express your answer to three significant figure and include the appropriate units. Submit Part B How many grams of ion at -10.3 degC can be completely to liquid at 21.0 degC 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 degC) and deltaH_max = 6.01 kJ/mol. Express your answer to three significant figures and include the appropriate units. Submit

Explanation / Answer

Calculate the enthalphy change for the process in which 37.7 g of water

Temperature change = 25.0-19.9= 5.1

The total enthalpy is include two process:

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

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.