Question

10g of ice at -20degree is added to 10g of water at 50 degree.the amount of ice and water that are present at equilibrium respectively.

Answer 1

Answer:

amount of water present = 15 g & amount of ice present = 5 g

Answer 2

Answer:

5g and 15g

Explanation:

Heat lost by ice = heat gained by water

m×s(specific heat)×∆t + m×L( latent heat of fusion) = m(of water) ×s∆t

m×1{0-(-20)} + m×80 = 10×1×50

20m+80m= 500

100m = 500

m( mass of ice melted) = 5g

Remaining mass of ice =( 10-5)g = 5g

Water in the mixture= 5g ( melted ice which is converted to water ) + 10g ( initial mass of water in the system) = 15g

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