Question

steam at 100 degrees ceicius is passed into a calorimeter of mass 55g, containing 150g of a mixture of ice and water, until all the ice has just melted. The mass of the calorimeter and its contents is found to be 217.5g.how much ice was originally present in the mixture? Given that the latent of fusion of ice =336,000J/Kg,latent heat of steam =2,270,000J/kg,specific capacity of water =1480J/kg.​

Answer 1

Hi it is 0.130
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Answer 2

Let M be the mass of steam condensed.

Amount of heat lost by the steam due to latent heat of condensation and due to cooling to 80

∘

C

=M×540+m(100−80)

Amount of heat absorbed by the water and calorimeter system=1100×1×(80−15)+20×1×(80−15)

From conservation of heat energy,

=M×540+M(100−80) =1100×1×(80−15)+20×1×(80−15)

⟹M=130g=0.130kg

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