Asked by Jordan
DrBob222,HELP!
Two 20.0-g ice cubes at –10.0 °C are placed into 275 g of water at 25.0 °C. Assuming no energy is transferred to or from the surroundings, calculate the final temperature of the water after all the ice melts.
heat capacity h2o (S)=37.7 j/mol*k
heat capacity h2o (l)=75.3 j/mol*k
enthalpy of fusion of h2o= 6.01 kj/mol
could you elaberate more with this. im still really cofused! thanks!
Two 20.0-g ice cubes at –10.0 °C are placed into 275 g of water at 25.0 °C. Assuming no energy is transferred to or from the surroundings, calculate the final temperature of the water after all the ice melts.
heat capacity h2o (S)=37.7 j/mol*k
heat capacity h2o (l)=75.3 j/mol*k
enthalpy of fusion of h2o= 6.01 kj/mol
could you elaberate more with this. im still really cofused! thanks!
Answers
Answered by
DrBob222
Substitute those numbers into the equation I gave you and solve for Tfinal. That is the only unknown in the equation; you will have numbers for everything else.
For example the first part was
mass ice x specific heat ice x (Tfinal-Tinitiaol).
mass ice = 40 g. Convert to mol(that'sw necessary because you have listed specific heat in J/mol*C). mol = grams/molar mass = 40/18 = 2.22 mol.
specific heat ice = 37.7 J/mol
Tfinal = unknown
Tinitial = -10
(2.22 x 37.7 x [Tf-(-10)] then do the next part etc.
It's nothing but arithmetic from here on out.
For example the first part was
mass ice x specific heat ice x (Tfinal-Tinitiaol).
mass ice = 40 g. Convert to mol(that'sw necessary because you have listed specific heat in J/mol*C). mol = grams/molar mass = 40/18 = 2.22 mol.
specific heat ice = 37.7 J/mol
Tfinal = unknown
Tinitial = -10
(2.22 x 37.7 x [Tf-(-10)] then do the next part etc.
It's nothing but arithmetic from here on out.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.