An ice cube of mass 0.500 kg and temperature –4.00° C is placed into 1.00 L of w

Question

An ice cube of mass 0.500 kg and temperature –4.00° C is placed into 1.00 L of water which is at 20° C. a. Determine the final temperature and state of the mixture. b. Starting from the state and temperature of the mixture found in part (a), how much energy needs to be added to the system by heat to fully convert the mixture into water vapor at 100° C? c. Starting from the state and temperature of the mixture found in part (a), how much energy needs to be extracted by heat to fully convert the mixture into ice at 0.00° C?

Explanation / Answer

Latent heat of fusion of water, L = 334000 J/kg
Latent heat of vapourisation of water, Lv = 2264760 J/Kg
Heat capacity of water, C = 4179 J/kg

m = 0.5 kg, t1 = - 4
M = 1 kg, t2 = 20

a. Assume that x mass of ice becomes watye where ass the water drops to temperature 0. If x comes out to be positive, than this assumption would be valid
Heat lost due to cooling water, H1 = MC(20 - 0) = 1*4179(20) = 83580 J
Heat provided by meting x amount of ice = xL = 334000x
Heat provided by bringing the ice to 0 degree = mC'(4) = 0.5*2030*4 = 4060 J
334000x + 4060 = 83580
x = 0.23808 kj ice melted
final M = 1.23080 Kg
final m = 0.26191 Kg
final T = 0
b. Heat required to change mixture into steam = 0.26191*334000 + 1.5*4179*100 + 1.5*2264760 = 4222467.94 J
c. Heat to be extracted = 1.23080*334000 = 411087.2 J

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