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.)

Question

1)54.2 kJ
2)33.3 kJ
3)54,220 kJ
4)20.9 kJ

Answer 1

1)54.2 kJ
2)33.3 kJ
3)54,220 kJ
4)20.9 kJ

Those are possible answers... Which is it?

Answer 2

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.

Answer 3

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

Answer 4

right. In fact, I obtained 54.2 kJ.