A 10.0 g piece of ice at 0 degree C is added to 20.0 g of water at 50.0 degree C
ID: 1061766 • Letter: A
Question
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 = 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. How would you best define the system and surroundings for this problem? What is the final temperature of the water? How could this process be done reversibly? If it were done reversibly, what would the values of and be? For the direct addition of ice to the warm wafer what are and?Explanation / Answer
So first thing in this mixing process will occur, is heat transfer from water at 50°c to saturated ice at 0°c till thermodynamic equilibrium occurs.
Since ice is saturated, so it will absorb heat in the form of latent heat to convert into water.
Let final temp.( equilibrium) of mixture is = T°c
So energy balance equation,
Latent heat to ice at 0°c to convert into saturated water at 0°c + sensible heating of this saturated water from 0°c to final temp T°c = cooling of hot water from 50°c to final temp T°c
Heat of fusion = 334.16 J g¯1
=> (10g*334.16 J g¯1) + {10g*4.184J/g.K*(T-0)K} = {20g*4.184J/g.K*(50-T)K}
=> (3341.6 + 41.84T)J = 83.68 (50-T) J
=> 3341.6 +41.84 T = 4184 - 83.68T
=> 125.52T = 842.4
=> T = 6.710C
The mixing is adiabatic and there is no heat transfer between the system and the surroundings.
This process could be made reversible by adding infinitismly small amounts of ice to water
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