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November 26, 2014

November 26, 2014

Posted by **Sarah** on Tuesday, October 23, 2012 at 2:21am.

- CHEMISTRY -
**Jennifer**, Tuesday, October 23, 2012 at 4:04pmthe heat capacities are given in units of moles and Kelvins, so you'll have to convert everything to these units

The weight of one mole of H2O = 1.008*2 + 15.999 = 18.015 g

17.68 g ice = 17.68g * (1 m / 18.015 g ) = .981 moles ice = .981 mole H20

54.05 g water = 3 moles liquid water

T(Kelvin) = T(Celsius) + 273

-12.7 C = 260.3 K

100 C = 373 K

First, the ice is brought to its melting point of 273 K:

37.5 J K-1 mol-1*.981 mol = 36.79 J K-1

The entropy change for this process is 36.79 ln (Tf - Ti)

where Tf is final temperature (273 K); Ti is initial temperature (260.3 K)

= 36.79 ln(12.7) = 93.5 J/K

Next, the ice melts:

.981 * 6.01 kJ/mol = 5.89 kJ

for a change of state at constant temperature the entropy change is

Q/T = 5.89 kJ / 273 K = 21.6 J/K

Next 17.68 g of ice at 273 K and 54.05 g water reach the same temperature; Using the first law of thermodynamics, deltaQ for this entire process is zero because the container is insulated. Summing up the heats of all the processes to 0:

36.79 J K-1 (273-260.3) + 5890 + .981 *75.3*(Tf-273) + 3*75.3*(Tf - 373) = 0

467.23 + 5890 + 73.87*(Tf-273) + 225.9*(Tf-373) = 0

6356 + 73.87*Tf - 20166 + 225.9*Tf - 84261 = 0

6356 +299.77*Tf - 104427 = 0

299.77*Tf = 98071

Tf = 327 K This is the final temperature of the mixture after it has reached equilibrium.

The change in entropy in bringing the melted ice to this temperature is

.981*75.3*ln(327-273) = 294.6 J/K

The change in entropy in bringing the liquid water down to this temperature is

3*75.3*ln(327-373) = an imaginary number

The total change in entropy is 93.5 J/K + 21.6 J/K + 294.6 J/K = 409.4 J/K

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