1) Calculate the enthalpy change, H , for the process in which 16.7 g of water i

ID: 475416 • Letter: 1

Question

1) Calculate the enthalpy change, H, for the process in which 16.7 g of water is converted from liquid at 13.3 C to vapor at 25.0 C .

For water, Hvap = 44.0 kJ/mol at 25.0 C and Cs = 4.18 J/(gC) for H2O(l).

Express your answer to three significant figures and include the appropriate units.

How many grams of ice at -5.2 C can be completely converted to liquid at 20.3 C if the available heat for this process is 4.63×103 kJ ?

For ice, use a specific heat of 2.01 J/(gC) and Hfus=6.01kJ/mol.

Express your answer to three significant figures and include the appropriate units.

delta H = H1 + H2

H1 = liquid from 13.3 C to 25C

H1 = m*Cliq*dT = 16.7*4.18*[25-13.3 ] = 816.73 J

H2 = Liquid at 25C to vapour

H2 = 16.7 *44 KJ / 18 = 40.822 KJ = 40822 J

delta H = 40822 + 816.73 = 41638.95 J = 41.638 KJ

Let the mass be 'm' grams of H2O

From ice at -5.2 C to 0C

Q1 = m*2.01*5.1 = 10.452 m

Q2 = ice at 0c to water

Q2 = 6.01KJ /mol

implies 6.01 KJ for 18 grams of H2O

For , m grams = 6.01*1000*m / 18 = 333.88 m

Q3 = Water at 0c to water at 23.C

Q3 = m*4.18*[20.3-0] = 84.854 m

4.63*10^3 *10^3 = 84.854m + 333.88 m+10.452 m

m = 10787.86 gms

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