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Topic: H and S  (Read 4535 times)

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Offline Dolphinsiu

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H and S
« on: October 08, 2007, 03:22:21 AM »
In a insulated flask (Dewar flask), 20 g of ice at -5C are added to 30 g of water at +25C. The pressure is kept constant throughout. Given the specific heat capacities are 4.18 JK-1g-1 and 2.09 JK-1g-1 for water and ice respectively, and the heat of fusion is 334 J/g. What is the final state of the system? Calculate Delta H and Delta S for the transformation? Is the process spontaneous?

20 g H2O (s,-5C) + 30 g H2O (l,25C) ---> (20 - x) g H2O (s,-5C) + (30 + x) g H2O (l,25C)

Isn't it?

x (5)(2.09) + x (334) + x (25) (4.18) = 448.95 x

But how can I know the mass of ice melted?

Also Delta G = H - T Delta S? What temperature need to be taken?

Offline Borek

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Re: H and S
« Reply #1 on: October 08, 2007, 04:10:10 AM »
heat gained = heat loss

In this case most likely

ice heating + ice melting + water from ice heating = water cooling

In general you should do it stepwise, starting with:

ice heating = water cooling

and checking after each step if the system is not already at equilibrium.

Note that ice melting heat is (mass melted)x(specific melting heat) and it occurs at constat temperature - that gives you enough information to calculate mass of ice melted; one equation with one unknown.
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Offline Dolphinsiu

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Re: H and S
« Reply #2 on: October 08, 2007, 11:10:07 AM »
20 (5) (2.09) + m (334) + 20 (25) (4.18) = 30 (25) (4.18)
m=2.50 g

Is that right?

Also is H = 2.5(334) = 835 J
         S = 835/273 = 3.06 J/K

G = H - TS = 835 - 273(3.06) = 0. Do I calculate it wrongly?

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