4. (20 points) A 10.0 g piece of ice at 0 degree C is added to 20.0 g of water a

Question

4. (20 points) A 10.0 g piece of ice at 0 degree C is added to 20.0 g of water at 50.0 degree C in a Dewar Flask. For water deltafus h^phi = 5980. J/mol. The heat capacity of liquid water may be considered constant for this question and is 4.184 J/gK. Consider the Dewar flask to be perfectly insulating with a negligible heat capacity. A. How would you best define the system and surroundings for this problem? B. What is the final temperature of the water? C. How could this process be done reversibly? 1f it were done reversibly. what would the values of deltaSsys and deltaSsurr be? D. For the direct addition of ice to the warm water what are deltaSsys and deltaSsurr?

Explanation / Answer

b)
Let us say that the final temperature of mixture of 10 grams of water at 0 degrees C, and 20 grams of water at 50 degrees C is t.

Then:

Heat gained by water at 0 Degrees C

= (Weight of water)*(Final temperature - Initial temperature)

= 10*(t - 0) = 10*t

Similarly:

Heat lost by water at 50 Degrees C

= (Weight of water)*(Initial temperature - Final temperature)

= 20*(50 - t) = 1000 - 20*t

But heat lost is equal to heat gained. Therefore:

10t = 1000 - 20*t

Therefor: 30t = 1000

Therefor: t = 1000/30 = 33.33 degrees C

Answer:

Final temperature is 33.33 degrees C.

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