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.
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!
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