Question

In a container of negligible mass 30g of steam at 100^@C is added to 200g of water that has a temperature of 40^@C If no heat is lost to the surroundings, what is the final temperature of the system? Also find masses of water and steam in equilibrium. Take L_v = 539 cal//g and c_(water) = 1 cal//g-^@C.

Answer 1

The final temperature of the system would be 100°C.The masses of water and steam in equillibrium is 222.26g and 7.74g

• Let us go through the given parameters in the problem

        Mass of steam(m1)=30g

       Mass of water(m2)=200g

       Temperature of steam (T1)=100°C

       Temperature of water (T2)=40°C

• Q be the heat required to convert 200 g of water at 40°C Into 100°C

              Q=mcΔT

               =(200)(1.0)(100−40)

               =12000cal

• The required heat be produced by conversion of m mass of steam into water then,

           m=(Q/L) = 12000/539 = 22.6 g

• As it is less than the mass of steam,still steam would be present in the mixture and the Temperature should be 100°C

• Mass of steam in the mixture =30−22.26=7.74g

• Mass of water in the mixture =200+22.26=222.26g