t Print Calculator Periodic Table Question 2 of 9 University Science Books Gener

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t Print Calculator Periodic Table Question 2 of 9 University Science Books General Chemistry 4th Edition presented by Sapling Learming Two 20.0-g ice cubes at -19.0 C are placed into 295 g of water at 25.0 oC. 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 H20(s) 37.7 J/(mol- K) heat capacity of H20 75.3 J/(mol. K) enthalpy of fusion of H2O 6.01 kJlmol no Previous ® Give up view salvtion 2 checks Answer Next Exit

Explanation / Answer

moles of ice = 40 / 18 = 2.22 mol

Energy required to raise ice from -19 to 0 degrees Celsius:

Q = n Cp t

Q = 2.22 x 37.7 x (0 - (-19))

Q1 = 1590.9 J

Energy required to completely melt the ice:

Heat of fusion for water: 6.01 kj / mol.

Q2 = Hf x n

Q2= 6.01 x 2.22 x 10^3

Q2 = 13348 J

now you need to calculate the energy to heat the 40 grams of cold water to the same temperature as the 295 grams of water, which is 25 degrees

(2.22mol)(-75.3J/mol)(25) = -4180J or -4.18kJ

now add up all those values,

- (1590.9 + 13348 + 4180 )= - 19119.9 J

   now add the masses together

295+40 = 335 grams of H2O,

moles = 335/18 =18.61 mols H2O

and finally

-19119.9 =(18.61)(75.3)(Tf-25)

Tf = 11.4 oC

final temperature = 11.4 oC

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