Question
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.)
1)54.2 kJ
2)33.3 kJ
3)54,220 kJ
4)20.9 kJ
1)54.2 kJ
2)33.3 kJ
3)54,220 kJ
4)20.9 kJ
Answers
Robert
1)54.2 kJ
2)33.3 kJ
3)54,220 kJ
4)20.9 kJ
Those are possible answers... Which is it?
2)33.3 kJ
3)54,220 kJ
4)20.9 kJ
Those are possible answers... Which is it?
DrBob222
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.
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.
Robert
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
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
DrBob222
right. In fact, I obtained 54.2 kJ.