Chem 201 . General Chemistry CH95HW Question 6 previous 1 6 of 20 n The constant

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Chem 201 . General Chemistry CH95HW Question 6 previous 1 6 of 20 n The constants for Hiare shown here Question 6 . Specific heat of ice 2.09J/( c . Specific heat of Iqad water svalr-418J/k . °C) Consider heating solid water (ice) until it becomes liquid and then gas (steam) (Egure.1) Altematively consider the everse process, cooling steam until ia becomes waber and inaly, ice (Eare 2 In each case, two types of transitions occur, those involringa temperature change with no change in phase (shown by the dagonal ine segments on the graghs) and those at constant temperature with a change in phase (shown by horirontal ine segments on the graphs Part A How mach heat energy, in kijouens, s negaiedt et of ke-1 to water t 25 C7 Express your answer to throe signilicent figures and include the appropriste units at-180 C to water at 250 C? Heating curve Value Units Hints My Answers Give Up Review Par Liquid and gas 100 Part B 6.3/s How long would tako for 1 50 mol of-ter at100 o-C to be converted completely-o steam if heat were added at a constriate of Express your answer to ee significant figures and include the appropriate units. Liquid Soles Hed added Value Unts Units Cortana. Ask me anything

Explanation / Answer

mass of water = 68 g

moles of water = 68 / 18 = 3.78

-18.0 °C ----------> 0.0 °C ------------> 25.0 °C

                 1 2 3

in this process 3 conversions are required.

Q1 = m Cp dT = 68 x 2.09 x (0 + 18) = 2558.16 J

Q2 = n x delta H fusion = 3.78 x 6.01 x 10^3 = 22717.8 J

Q3 = m Cp dT = 68 x 4.18 x (25 - 0) = 7106 J

total heat = Q1 + Q2 + Q3

= 32.4 kJ

heat energy = 32.4 kJ

part B)

moles of water = 1 .50

mass of water = 1.50 x 18 = 27 g

Q = n x delta H = 27 x 2250 = 60750 J

time = 60750 / 16

time   = 3.80 x 10^3 s

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