Question

Part A) Calculate the enthalpy change, ΔH, for the process in which 38.6 g of water is converted from liquid at 0.3 ∘C to vapor at 25.0 ∘C .

For water, ΔHvap = 44.0 kJ/mol at 25.0 ∘C and Cs = 4.18  J/(g⋅∘C) for H2O(l).

Part B) How many grams of ice at -24.6 ∘C can be completely converted to liquid at 9.4 ∘C if the available heat for this process is 4.04×10³ kJ ?

For ice, use a specific heat of 2.01 J/(g⋅∘C) and ΔHfus=6.01kJ/mol

Answer 1

Three are three changes will occur to the ice:

1) heat from -24.6 to 0
2) melt at 0
3) heat from 0 to 9.4

The mass is the same value in each situation.

q1 = (x) (24.6) (2.01 J/(g. degC))
q2 = (x / 18.015 g/mol) (6010 J/mol)
q3 = (x) (9.4) (4.184 J/(g. degC))

q1+q2+q3 =  4.04×103

(x) (24.6) (2.01) + (x / 18.015 g/mol) (6010 J/mol) + (x) (9.4) (4.184 J/(g. degC)) =  4.04×10⁶ Joules

49.446 X + 333.888 X + 39.3296 x =  4.04×10⁶ Joules

422.6644 X =  4.04×10⁶ Joules

X = 9558 gm of ice

Hence 9558 gm or 9.558 kg of ice can be converted from  -24.6 ∘C to 9.4 ∘C