Question
The heat of fusion of water is 6.01 kJ/mol. The heat capacity of liquid water is 75.3 J/mol • K. The conversion of 50.0 g of ice at 0.00°C to liquid water at 22.0°C requires how much kJ of heat?
Can someone explain how to do this, I tried but keep getting the wrong answer.
Can someone explain how to do this, I tried but keep getting the wrong answer.
Answers
DrBob222
50/18.015 = moles in 50 g H2O.
q1 = heat to melt ice.
q1 = moles H2O x 6.01 kJ/mol
q2 = heat to move ice water from zero C to 22 C.
q2 = moles water x 75.3 J/mole x (22-0).
Change q1 to J and add to q2.
q1 = heat to melt ice.
q1 = moles H2O x 6.01 kJ/mol
q2 = heat to move ice water from zero C to 22 C.
q2 = moles water x 75.3 J/mole x (22-0).
Change q1 to J and add to q2.
maria
10.6kj
adi
21.3 kJ of heat.
no explanation sorry im tkaing a test and i just wanted ot help i found out this was the anwser lmao
no explanation sorry im tkaing a test and i just wanted ot help i found out this was the anwser lmao