An ice cube of mass 0.5kg and temperature T=-4 degrees Celsius is put into 1 lit
ID: 867053 • Letter: A
Question
An ice cube of mass 0.5kg and temperature T=-4 degrees Celsius is put into 1 liter of water which is at room temperature (20 degrees Celsius).
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 heat needs to be added to the mixture to fully convert the mixture into vapor at 100C?
c. Starting from the state and temperature of the mixture found in part (a), how much heat needs to be extracted to fully convery the mixture into ice at 0C?
Explanation / Answer
(a). Here heat will be absorbed by ice cube and released by water at room temperature.
For ice: mass, m1 = 0.5 kg
Temperature T1 = - 4 DegC
For water: mass, m2 = 1 kg (since density of water is 1g/mL)
Temperature T2 = 20 DegC
Heat required by ice to melt completely
= 0.5kg x 2093Jkg-1DegC-1x[0 –(-4)] + 0.5kg x336000Jkg-1 =
= 4186J + 168000J = 172186 J
Heat given off by water when its temperature decreases to 0 DegC
= 1kg x4186Jkg-1DegC-1x20 DegC = 83720 J, which is less than 172186 J.
Since heat given off by water is less than the heat required, all ice will not be melted.
Hence final temperature will be 0 DegC and the mixture will contain both ice and water.
(b). Heat required by the mixture to fully convert to vapor at 100 DegC is
= Heat required by 1L water at 20 DegC + Heat required by 0.5 kg ice at -4 DegC
= [1kg x4186Jkg-1DegC-1x(100 – 20) DegC + 1kg x 2260000 Jkg-1] + [0.5kg x 2093Jkg-1DegC-1x[0 –(-4)] + 0.5kg x336000Jkg-1 + 0.5kg x4186Jkg-1DegC-1x(100 – 0) DegC + 0.5kg x 2260000 Jkg-1]
= 334880J + 2260000J + 4186J + 168000J + 209300J + 1130000J
= 4106366J = 4106 KJ (answer)
(c). Heat extracted from water to convert to ice at 0 DegC
= 1kg x4186Jkg-1DegC-1x(0 - 20) DegC – 1 kg x336000Jkg-1
= - 419720 J
Heat received by ice at – 4 DegC to convert to water at 0 DegC
= + 0.5kg x 2093Jkg-1DegC-1x[0 –(-4)] = + 4186J
Hence net heat extracted = - 419720 J + 4186J
= - 415534 J = - 415.5 kJ (answer)
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