Question

Finding the equilibrium temperature of a mixture: An isolated thermal system consists of a copper container filled with a quantity of liquid water and a quantity of ice. What is the fully thermalized state of the system (the final temperature, how much water, and how much ice) provided that initially there is 1.0 kg of ice at -100 degrees Celsius, 10 kg of water at 1 degrees Celsius, and the copper container has the mass of 15.0 kg and is initially at 10 degrees Celsius? The heat capacities of ice, water, and copper are 2050, 4186, and 385 J/degrees Celsius, and the latent heat of ice is 3.33 x 10^5 J/kg.

Answer 1

let the final temperature be 0 degree celcius such that water and ice coexist

let mass of water converted to ice be m kg.

then heat released =heat released by copper + heat released by water while reducing temperature + heat released by m kg of water while converting to ice

=15*385*(10-0)+10*4186*(1-0)+m*3.33*10^5

=99610+m*3.33*10^5 J

this heat is used to heat the ice from -100 degree celcius to 0 degree celcius

heat required to do so=1*2050*(0-(-100))=205000 J

equating heat released to heat absorbed:

99610+3.3*10^5*m=205000

==>m=(205000-99610)/(3.3*10^5)=0.31936 kg

hence 0.31936 kg of water will become ice.

so final temperature is 0 degree cecius

mass of ice=1+0.31936=1.31936 kg

mass of water=10-0.31936=9.6806 kg