o 12/8/2017 11:00 PM 086.7/1001 2/8/2017 04:39 PM Gradebook -Print Calculator -a

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o 12/8/2017 11:00 PM 086.7/1001 2/8/2017 04:39 PM Gradebook -Print Calculator -all Periodic Table Question 22 of 22 Incorrect University Science Books presented by Sapling Learning Tutorial leson Map fh General Chemistry 4th Edition o 200-g ice cubes at-19.0 are placed into 225 g of water at 25.0 oC. Assuming no energy is transferred to or from the surroundings, calculate the final temperature, Tf, 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 H20 6.01 kJ/mol Number 7, = 11 15.7 o C Previous Give Up & View Solution # Try Again Next Exit

Explanation / Answer

Answer

Tf = 7.75

Explanation

Heat gained by ice cubes(qice) = - Heat lost by liquid water(qwater)

qice = Heat required to raise from -19 to 0(q1) + Heat required to melt ice(q2) + Heat required to raise the temperature of melted ice from 0 to Final temperature X

q1 = m × T × C

= 40g × 19K × 2.093J/g K

= 1591J

q2 =( 6010J/18.014g)× 40g = 13345J

q3 = m × T × C

= 40g × (X - 273.15K) × 4.184J/g K

= 167.36J/K - 45714J

qice = 1591J + 13345J + 167.36J/K - 45714J

= -30778J+ X167.36J/K

qwater = m × T × C

= 225g × (X - 298.15K) × 4.184J/g K

= X941.4J/K - 280678J

So,

-30778J + X167.36J/K =- (X941.4J/K - 280678J/K)

X1108.76J/K = 311456J

X = 280.90K = 7.75

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