Answer to Question #68713 in Molecular Physics | Thermodynamics for ujjwal
What happens when 5 g of ice at -20c is added to 5 g of water at 30c. (note:all ice has not changed into water)
The water transfers heat to ice thus heating it and making it melt. Let x represent the mass of the ice that will melt. The heat supplied by the water cooling down is (4200 J/kgK) × (0.005 kg) × (30 K) The heat received by the ice is (2050 J/kgK) × (0.005 kg) × (20 K) + (334 KJ/kg) × (x kg) Assuming the system isolated, the heat lost by the warm water is equal to the heat received by the ice. 4200 x 0,005 x 30 = 2050 x 0.005 x 20 + 334.000x Solving for x, obtaining x = 0.00127. Therefore, 1.27 g of ice will melt, and the total mass of 8.73 g of water will remain at 0°C.