Question

Mass of Styrofoam cups	4.00 g
Mass of Styrofoam cups with water	102.30 g
Mass of Water	98.3
Initial Temperature of water	22.0  oC
Final Temperature	11.4  oC
Mass of Styrofoam cups with water plus ice	113.60 g
Mass of Iced Added	11.3 g

1. Ice at -10 °C is heated until it becomes liquid water at 10°C. Which heating curve below represents this process? 10| و (2) T T ( °C او ( T (°C) 0 -10 سے ان کا -10 ان heat heat heat (a) (b) (c) Ans. 2. Convert the heat of fusion of water, 334 J/g, to kJ/mol. Show your calculation. Ans. 3. What mass of water at 25.0°C will just melt 50.0 g of ice at 0.0°C when they are mixed? Show your calculation.

Answer 1

1. heating fusion heating -10°c ice oʻC ice o c water 10°C water overing fusion (mellings, tomperature remains constant. 10°C | Answer. Option Temperature oc -10C Heat t

2. Given heat of farion = 334 I of water hear of fusion - 334 X10-9KJ [13= 10-3 ki] Leve kusw mass of 1 mole of water = 18g le. 18g waren = 1 mole waren So, Ig water = 1/3 mole wath pulting this value in heat of fusion- heat of fusion = 334 x 10-3 kJ Ga mor) - 334x10-3 x 18 kJ hreat of fusion of water = 6.012 KJ/mol moy 3. As per law of conservation of energy, heat lost by water at as a WU be equal to heat absorbed by sog ice at 0 °C for fusion. Also, as per law of thermodynamics", temperature of hot water will finally become o°C of equilibrium So, heat tout by water = heat gained by ice during fusion. /MC or l = m. All fusion * mx 4. 186 J/gt og x 25% = 50gx 334,8g um = 50x334 g = 159.589 mass of water required = 159.589