An insulated container holds 1.00 L of warm water at an initial temperature of 3

Question

An insulated container holds 1.00 L of warm water at an initial temperature of 37.0°C. Then 825 g of ice, initially at 0°C, is added to the water. The ice begins to melt as the water cools, and eventually equilibrium is reached with both ice and water present at 0°C. What mass of ice remains unmelted?

Densities: Ice = 0.917 g/cm3

Liquid water = 1.00 g/mL

Heat capacities: Ice = 37.5 J/mol/K = 2.08 J/g/°C

Liquid water: = 75.3 J /mol/K = 4.18 J/g/°C

Heat of fusion of ice = 6.01 kJ/mol

Heat of vaporization of water = 40.7 kJ/mol

**Answer should be 361 g, but how??

Explanation / Answer

Warm water quantity= 1 L =1000ml

Mass of water= 1000*1= 1000 gm

Since the container is insulted, there is no heat loss to the surroundings

Heat lost by hot water= Sensible heat= mass*specific heat* temperature difference = 1000* 4.18*( 37-0)= 154660 joules

Let x be the amount of ice that is melted. In the process of melting, ice gains heat of fusion.

x* 6.01Kj/mol= x*6.01*1000/18=334 joules= 154660,x =463

ice that is unmelted= total ice- ice that is melted= 825-463= 362 gms

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