Question

In the previous question you were provided with an extra piece of information: you were told that only liquid water would be present once equilibrium was established. if the final state of a system is not given, it is still possible to determine it from the initial temperatures and masses. the next few parts will help illustrate this point. suppose that in an insulated container, 0.100 kg of water at 20.0∘c is mixed with 1.500kg of ice at −15.0∘c. you are asked to find the final temperature tf of the system, but you are not told what the final phase of the system in equilibrium is.

Answer 1

Answer:   
 Latent heat fusion of water, hf = 334 kJ/kg 
 specific heat of water is 4.186 kJ·kgâ’1·Kâ’1 
 specific heat of ice is 2.05 kJ/kgK 
  
 First, lets assume the ice will melt and become liquid water 
 Q in = Q out 
 m1*C(ice)*(0 deg C- (-) 15) + m1*hf + m1*C(water) (Tf - 0))= m2*C(water)*
(T2 - Tf) 
 1.5* 2.05*15 + 1.5*334 + 1.5*4.186*Tf = 0.1*4.186 (20 - Tf) 
 46.125 + 501 + 6.279 Tf = 0.4186 (20 - Tf)... (1)   
 For water if it decrease to 0 deg C, the energy as heat release is 0.4186 (20
- 0) = 8.372 kJ < 46.125 kJ, Lower than Ice which increase temperature from
-10 to zero. So Water will become ice.   
 The energy dissipate for 20 deg C water and change its phase to 0 deg C ice
is still lower than energy absorb by ice to 0 deg C, rearrange the equation.   
 Q in = Q out 
 m1*C(ice)*(Tf- (-) 15) = m2*C(water)*(T2 - 0) + m2*hf + m2*c(ice) (0-Tf) 
 1.5* 2.05* (Tf +15) = 0.4186*20 + 0.1*334 + 0.1*2.05 (-Tf) 
 3.075 (Tf + 15) = 8.372 + 33.4 + 0.205 (-Tf) 
 3.075Tf + 46.125 = 41.772 - 0.205 Tf 
 3.28 Tf = - 4.353 
 Tf = - 1.327 deg C ( Ice phase)