Question

The water equivalent of a calorimeter is 100g. It contains 50g of water at 20°C. How much of dry ice
at 0°C should be added to it to reduce the temperature of water to 10°C? [specific heat of ice
= 0.5 calg loc latent heat of fusion of ice = 80 calg ]​

Answer 1

Answer:

66.66 g of dry ice is required.

Explanation:

One of the container contains 50 g water at 20°C

Let the amount of dry ice at 0°C should added to it to reduce the temperature of water to 10°C

Therefore the amount of dry ice will be,

m x 1/2 x 10 + m/2 x 80 = 150 x 1 x 20

or, 45 m= 3000

m= 3000/45

m= 66.66