Question

A Copper calorimeter of mass 100gm contains 200gm of a mixture of Ice and water. Steam at 100° under normal pressure is passed into the calorimeter and the temperature of the mixture is allowed to rise to 50°C. If the mass of the calorimeter and its content is now 330gm, What was the ratio of the ice and water in the begining? Neglect heat losses.

Answer 1

We can solve it simply by:

Check the units conversion correctly if it doesn't matches.

Answer 2

Mass of calorimeter = 100g= 0.1kg

Let the mass of ice in the calorimeter = x kg

then initial mass of water in the calorimeter =(0.2-x)kg

Total mass of water at 0^oC after the ice melts =0.2kg

As the final mass of the calorimeter with its content is 0.33kg then the mass of steam condensed will be =0.03kg.

Heat lost by the steam due to condensation  as water at 100^oC is = 0.03*22.5*10^5 J

Heat lost by 0.03 kg water of 100^oC to reach at final temperature 50^oC will be

 = 0.03*4.2*10^3 *50J

Heat gained by x kg ice to melt = x*3.36*10^5 J

Heat gained by water of mass 0.2*kg to reach at 50^oC from 0^oC = 0.2*4.2*10^3*50 J

Heat gained by calorimeter = 0.1*0.42*10^3*50 J

By calorimetric principle

x*3.36*10^5+0.2*4.2*10^3*50+ 0.1*0.42*10^3*50=0.03*22.5*10^5+ 0.03*4.2*10^3 *50

=>3.36x+0.2*2.1+0.021=0.675+0.063

=>1.6x+0.2-+0.01=0.675/2.1+0.03

=>1.6x=0.3514+0.02-0.2

=> x=0.1414/1.6= 0.0884kg

Mass of ice = 0.0884kg and mass of water = 0.2-0.0884=0.1116g

So the ratio of ice and water in the mixture = 884.:1116