Asked by Jhon
How much energy is needed to heat 8.50 g of ice from -30.0 C to 70.0 C?
The heat of fusion of water is 6.01 kJ/mol, and the molar heat capacity is 36.6 J/(mol.K) for ice and 75.3 J/(mol.K) for liquid water.
I got 19.24kJ but it isn't correct!!
The heat of fusion of water is 6.01 kJ/mol, and the molar heat capacity is 36.6 J/(mol.K) for ice and 75.3 J/(mol.K) for liquid water.
I got 19.24kJ but it isn't correct!!
Answers
Answered by
DrBob222
You do this in steps.
q1 = energy needed to heat solid ice from -30 C to solid ice at 0 C.
q1 = mass ice x specific heat ice x (Tfinal-Tinitial) where Tf is 0 and Ti is -30 C.
q2 = heat needed to change the phase at zero C from solid ice to liquid water.
q2 = mass ice x heat fusion
q3 = heat needed to raise the T of liquid water at zero C to 70 c.
q3 = mass water x specific heat liquid H2O x (Tfinal-Tinital) = ?
NOTE: Make sure the units are ok. If you intend to use J/mol K for specific heat then you MUST use mass in mols and not grams. My suspicion is this is where you went wrong.
Total is q1 + q2 + q3
q1 = energy needed to heat solid ice from -30 C to solid ice at 0 C.
q1 = mass ice x specific heat ice x (Tfinal-Tinitial) where Tf is 0 and Ti is -30 C.
q2 = heat needed to change the phase at zero C from solid ice to liquid water.
q2 = mass ice x heat fusion
q3 = heat needed to raise the T of liquid water at zero C to 70 c.
q3 = mass water x specific heat liquid H2O x (Tfinal-Tinital) = ?
NOTE: Make sure the units are ok. If you intend to use J/mol K for specific heat then you MUST use mass in mols and not grams. My suspicion is this is where you went wrong.
Total is q1 + q2 + q3
Answered by
HARRYAZCRACK
Chemistry sucks dog nuts
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