How do I determine the questions with the calculations I already have? mass of c

ID: 909988 • Letter: H

Question

How do I determine the questions with the calculations I already have?

mass of coffee cup 3.86 g

mass of coffee cup and water 168.42 g

mass of water 164.55 g

moles of water ?

mass of coffee cup, water and steam 143.31 g

mass of steam -25.11 g

moles of steam ?

initial temperature 22.6 degrees celcius

final temperature 82.1 degrees celcius

delataT of water 59.5 degrees celcius

heat gained by water (qwater) (J) ?

DeltaH/mol water (kJ/mol) ?

heat lost by steam (qateam) (J) ?

DeltaT of Condensed steam (degrees celcius)?

heat lost by condensed steam(qcooling) (J) ?

heat needed to condense steam(qcondensaton) (J)?

DeltaHvaporization/mol steam (kJ/mol) ?

First weight of steam cannot be -ve

So I would take here a +ve value approximately same but opposite sign,

mass of steam = 25.11 g (If you think you have correct value, do the calculations with -ve value, but that shouldn't be right)

mass of water = 168.42 - 3.86 = 164.56 g

moles of water = g/molar mass = 164.56/18.02 = 9.132 mols

moles of steam = g/molar mass = 25.11/18.02 = 1.393 mols

(deltaT)water = Tf-Ti = 82.1 - 22.6 = 59.5 oC

(deltaT)steam = Ti-Tf = 100 - 82.1 = 17.9 oC

Heat gained by water = 164.56 x 4.184 x 59.5 = 40967 J

deltaH/mol of water (kJ/mol) = 40967/9.132 = 4.49 kJ/mol

heat lost by steam = -40967 J

deltaT of condensed steam = 17.9 oC

heat lost by condensed steam = 40700 x 1.393 = 56695 J

heat needed to condense steam (qcondensation) = 56695 J

deltaHvap steam (kJ/mol) = 56697/1.393 = 40.700 kJ/mol

heat required to condense steam + heat required to warm cold water
40700 x 1.393 + 25.11 x 4.184 x 17.9

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