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A 100 gram calorimeter with a specific heat of 0.20 cal/ g-c contains a mixture of ice and 200 grams water. when 40 grams of steam is added to the mixture, the temperature is raised to 60 celsius. how many grams of ice were originally in the calorimeter?

Shubhangi Shilendra
30 Points
2 years ago

Heat gained by calorimeter+ heat gained by water to reach 60 degrees + heat gained by ice to melt to water
at 0degrees + heat gained by melted ice to reach 60degrees= heat provided by steam to convert into water+
heat lost by condensed steam at 100degrees to reach 60 degrees

Let mass of ice present in mixture be ‘a’ grams.
Therefore,
(100x0.2x60)+( 200x1x60)+ (a x 80)+ (ax1x60)= (40x540) +( 40x1x (100-60))
1200+12000+140a= 21600+1600
13200+140a= 23200
140a=10000
a=1000/14=71.42g
i hope this is the right answer. thanks for the question :D