Question

Calculate the final equilibrium temperature when 10.0 grams of steam initially at 100 degree C is mixed with 450 grams of liquid water and 110 grams of ice at 0 degree C in a calorimeter. That is, the liquid water AND the ice are initially at 0 degree C. Ignore any heat energy exchanges with the calorimeter and the surroundings. If you conclude that the final temperature of the system is 0 degree C, then what mass of ice remains, BUT if you conclude that the final temperature is 100 degree C, then what mass of steam remains? [For water: the latent the latent heat of fusion is 3.33 times 10^5 J/kg, the specific heat capacity is c_ water = 4186 J/(kg degree C), and the latent heat of vaporization is 2.26 times 10^6 J/kg).

Answer 1

(a) The principle of calorimetric states as follows: lloss = QGAIN By considering the final temperature of the mixture as 0°C. The heat liberated by water vapor is given by the specific heat equation as follows: Q = mL, + mcwote, 100°C Substituting we get, Q =([(0.01 kg)(2.26x10$J/kg)]+[(0.01 kg)(4186J/kg• °C)(100°C)]) = 26786J This heat is given to the ice to convert to water. Q= The heat gained by ice is as follows: Q2 = mL; Substituting we get 26786J = m(3.33x10 J/kg) m = 0.080 kg The mass of ice remains is, m = 0.080 kg

(b) By considering the final temperature of the mixture as 100°C. Heat gained by ice is, Q = mL, + Mcwater (100°C) =(0.11kg) (3.33x10ʻJ/kg)+(0.11kg) (4186J/kg . °C)(100°C) = 826761 The heat gained by liquid water is as follows: Q =mcwaer (100°C) Substituting the values, we get, Q2 = (0.45 kg)(4186J/kg.°C)100°C =1883701 This heat is greater than that the heat energy produced by the total mass of the steam. Hence, the equilibriumtemperature as 100°C is not possible. No vapor will remain.