1. A 25g ice cube, when left outside on a summer day, will spontaneously melt. D

Question

1. A 25g ice cube, when left outside on a summer day, will spontaneously melt.  During this process, the temperatures of both the ice cube (0oC) and the surroundings (30oC) remain constant. Calculate the change in entropy (in J/K) for the UNIVERSE, given that the heat of fusion for water is 334 J/g.

2. What is the change in entropy, in J/K, when 150 g of argon (heat of vaporization = 6.43 kJ/ mol) condenses from a vapor to a liquid at its boiling point, - 186oC?

3. Solve for x:

-log 7 x = 1.6

(Note the "subscript" 7.)

Explanation / Answer

1. Change of enthalpy when 25 g of ice melts: 334 J/g * 25 g = 8350 J

So, entropy change of universe: Heat gained by universe/ temperature of surroundings = 8350 J/ ( 30+273) K = 8350 J/ 303 K = 27.56 J/K

2. Atomic mass of argon is : 40 g / mole

Moles of argon in 150 g = 150 g/ (40 g/mole) = 3.75 mole

Heat released when 3.75 mole of argon vapor is condensed to liquid:

- 3.75 mole * 6.43 kJ/ mole = - 24.1 kJ = - 24.1 * 103 J

Entropy change = heat absorbed/ boiling point = (-24.1 * 103 J )/ (-186 +273)K = - 277 J/K

3. - log7x = 1.6

Or, log7x = -1.6

Or, x = 7-1.6

Or, x = 0.044

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