sarahcanflyy
New member
Two 20.0 gram ice cubes at -11.0 degrees Celsius are placed into 275 grams of water at 25.0 degrees Celsius. Assuming no energy is transferred to or from the surroundings, calculate the final temperature of the water after all the ice melts.
heat capacity of H2O (s) = 37.7 J/molK
heat capacity of H2O (l) = 75.3 J/molK
enthalpy of fusion of H2O = 6.01 kJ/mol
I thought I did this right but I must have have made an error somewhere. If someone could point out that error and point me in the correct direction, it would be greatly appreciated.
melting point of water = 0 degrees Celsius = 273.15 K
-11.0 degrees Celsius = 262.15 K
40 g H2O (s)= 2.22 mol
275 g H2O (l) = 15.26 mol
qp = nCpdeltaT
qp = (2.22 mol)*(37.7 J/mol K)*(273.15 K - 262.15 K)
qp = 920.634 J
qp = ndeltaH
qp = (2.22 mol)*(6.01 kJ/mol)
qp = (13.34 kJ)*(1000 J)
qp = 13342.2 J
qp = 920.634 J + 13342.2 J
qp = 14262.834 J
deltaT = (qp)/(n*Cp)
delta T = (14262.834 J)/(15.26 mol*75.3 J/molK)
delta T = 12.41
T = 25 degrees Celsius - 12.41 degrees Celsius
T= 12.59 degrees Celsius
(40.0 g)*(Tf-0) = (275 g)*(12.59 degrees Celsius-Tf)
(40.0 g)*Tf = 3462.25 - (275 g)*Tf
(315 g)*Tf = 3462.25
Tf = 10.99 degrees Celsius (which is supposedly incorrect)
I did a tutorial problem of this problem with all of the same values but the ice cubes where at -10.0 degrees Celsius instead of -11.0 and I got 11.09 degrees Celsius as the correct answer so I know this problem has to be somewhere around 10 degrees Celsius but I'm not sure what I did wrong because I did everything the same as in the tutorial problem except for changing the temperature value. Any help is appreciated.
heat capacity of H2O (s) = 37.7 J/molK
heat capacity of H2O (l) = 75.3 J/molK
enthalpy of fusion of H2O = 6.01 kJ/mol
I thought I did this right but I must have have made an error somewhere. If someone could point out that error and point me in the correct direction, it would be greatly appreciated.
melting point of water = 0 degrees Celsius = 273.15 K
-11.0 degrees Celsius = 262.15 K
40 g H2O (s)= 2.22 mol
275 g H2O (l) = 15.26 mol
qp = nCpdeltaT
qp = (2.22 mol)*(37.7 J/mol K)*(273.15 K - 262.15 K)
qp = 920.634 J
qp = ndeltaH
qp = (2.22 mol)*(6.01 kJ/mol)
qp = (13.34 kJ)*(1000 J)
qp = 13342.2 J
qp = 920.634 J + 13342.2 J
qp = 14262.834 J
deltaT = (qp)/(n*Cp)
delta T = (14262.834 J)/(15.26 mol*75.3 J/molK)
delta T = 12.41
T = 25 degrees Celsius - 12.41 degrees Celsius
T= 12.59 degrees Celsius
(40.0 g)*(Tf-0) = (275 g)*(12.59 degrees Celsius-Tf)
(40.0 g)*Tf = 3462.25 - (275 g)*Tf
(315 g)*Tf = 3462.25
Tf = 10.99 degrees Celsius (which is supposedly incorrect)
I did a tutorial problem of this problem with all of the same values but the ice cubes where at -10.0 degrees Celsius instead of -11.0 and I got 11.09 degrees Celsius as the correct answer so I know this problem has to be somewhere around 10 degrees Celsius but I'm not sure what I did wrong because I did everything the same as in the tutorial problem except for changing the temperature value. Any help is appreciated.