25 kg of ice at 0° C is combined with 4 kg of steam at 100° C. What is the final equilibrium temperature in ° C of the system? (Lf = 3.33 x 105 J/kg; Lv 2.26 x 106 J/kg, specific heat capacity of water =4186 J/(kg.K))

Question

Question 2 options:
A) 40

B) 20

C) 60

D) 100

E) 8

Answer 1

If all the ice melts, 25*Lf = 8.33*10^6 kJ is absorbed in that process. If all of the steam condenses, 4*Lv = 9.04*10^6 kJ is released in that process. There is enough steam to melt all the ice.

0.71*10^6 J are still available from condensation of remaining to raise the temperature of the melted ice. To raise all of the 25 kg to 100 C would require 1.047*10^7 kJ. So the melted water will be raised to an intermediate temperature between 0 and 100 C.

Calculate the final temperature by assuming all of the steam condenses and is lowered to final temperature Tf, releasing heat that melts all the ice and raises it to the same Tf.

25*(3.33*10^5 + 4186 Tf) = 4*(2.26*10^6 +4*(100 - Tf)]

Solve for Tf

Answer 2

60