posted by Robert .
Calculate the enthalpy change when 100. g of ice at 0.0 °C is heated to liquid water at 50.0°C. (The heat of fusion for water is 333 J/g.)
Those are possible answers... Which is it?
q1 = heat to melt the ice.
q1 = mass of ice x heat of fusion.
q2 = heat to raise the temperature from 0 degrees C to 50 degrees C.
q2 = mass of water x specific heat water x (Tf - Ti)
Tf is final Temperature.
Ti is initial T.
Total heat = q1 + q2.
Post your work if you get stuck.
Never mind, I got it:
100 grams of ice times 333 J/g = 33.3 kJ of heat to melt all the ice.
Next, raising temperature of liquid, so 100 g *4.184 J/g *50= 20.92 kJ
Thus A is best answer
right. In fact, I obtained 54.2 kJ.