Calculating enthalpy change? Hess's Law?

Calculate the enthalpy change (in kJ) associated with the conversion of 25.0 grams of ice at -4.00 °C to water vapor at 109.0 °C. The specific heats of ice, water, and steam are 2.09 J/g•K, 4.18 J/g•K, and 1.84 J/g•K, respectively. For H2O, Δ Hfus = 6.01 kJ/mol and \DeltaHvap = 40.67 kJ/mol.

Please just post the answer for now, don't worry about the explanation until after I respond. Thanks

1 Answer
Al E.
Nov 21, 2017

#(i) + (ii) + (iii) + (iv) + (v) approx 75.5kJ#

Explanation:

#(i)# #25.0g * (2.09J)/(g*°C) * 4°C = 209J#
#(ii)# #25.0g * (H_2O)/(18.0g) * (6.01kJ)/(mol) approx 8.35kJ#
#(iii)# #25.0g * (4.18J)/(g*°C) * 100°C = 10450J#
#(iv)# #25.0g * (H_2O)/(18.0g) * (40.67kJ)/(mol) approx 56.49kJ#
#(v)# #25.0g * (1.84J)/(g*°C) * 9.0°C approx 5.11J#

Related questions
Impact of this question
5543 views around the world
You can reuse this answer
Creative Commons License