A 15-kg block of ice is cooled to -100 degree C and is then added to 8.00 kg of

Question

A 15-kg block of ice is cooled to -100 degree C and is then added to 8.00 kg of water in a 10-kg copper calorimeter at a temperature of 25 degree C. cw = 4186 J/kg degree C, ccu 387 J/kg degree C, cice = 2090 J/kg degree C, Lfice = 3.33 x 10^5 J/kg. Assume the system (ice, water, and copper) is isolated. a. Find the mass of the ice that was melted (if no ice melted, then report 0). b. Find the mass of the water that froze (if no water froze, then report 0). c. Determine the final temperature of the system. Prof e d. Calculate the change in entropy of the system.

Explanation / Answer

Heat lost by water and calorimeter as the temperature decreases from 25 to 0,

Q1 = m_water*c_water*(25- 0 ) + m_copper*c_copper*(25-0)

= 8*4186*25 + 10*387*25

= 933950 J

heat gained by ice as they temperature rises from -100 to 0,

Q2 = m_ice*c_ice*(100-0)

= 15*2090*100

= 3135000 J

clearly Q2 > Q1.

it means certain amount of water becomes ice

let m is the mass of water that becomes ice

Q2 = Q1 + m*Lf

==> m = (Q2-Q1)/Lf

= (3135000 - 933950)/(3.33*10^5)

= 6.6 kg

a) 0

b) 6.6 kg

c) 0 degrees centigrade

d) I don't know The formula

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