Question

0 points) A 10.0 g piece of ice at 0°C is added to 20.0 g of water at 50.0°C in a Dewar Flask. For water Δ'usHa_ 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?

Answer 1

For the given Dewar flask reaction mixture of ice at 0oC at water at 50oC when mixed together we could define the system and the surrounding as,

a. The system in this case is mixture of ice and water. The ice here would gain heat from the hot water to raise its temperature and gets melted into water. The heat of fusion is used for this calculation. Hot water looses the heat in the process to ice.

So,

heat lost by hot water = heat gained by ice to raise its temperature + heat gained by ice to melt it into water

heat = mCpdT

with,

m = mass of ice or water

Cp = specific heat capacity of ice or water

dT = change in temperature

For melting ice,

heat = mdHfus

with,

m = mass of ice

dHfus = heat of fusion

Therefore feeding all the values we could get final temperature (Tf) of the system,

20g x 4.184J/g.K x (50 - Tf)oC = 10g x 5980J/mol/18g/mol + 10g x 2.11J/g.K x (Tf - 0)

4184 - 83.68Tf = 3322.22 + 21.1Tf

Tf = 861.78/104.78 = 8.22oC

So the final system temperature would be 8.22 oC, which means all of ice has melted and now we have only water at 8.22 oC in the system. The surrounding here is insulated walls of the Dewar flask, which does not take up any heat from the system.