The effect of heat transfer on objects undergoing a phase transition is differen
ID: 3161886 • Letter: T
Question
The effect of heat transfer on objects undergoing a phase transition is different from its effect on objects undergoing a temperature change. Transfer of heat is still thought of as the cause of what happens, but the effect of the heat transfer is different than usually expected. Instead of raising the temperature of the ice, the heat melts the ice without changing its temperature. There is another difference as well, In interactions with no phase change, the heat received by the cold object is evenly distributed shared - equally among all of the grams (mass) of the cold substance. In the case of melting, all of the heat goes to just the ice that melts. No heat goes to the unmelted ice The following exercise is a "thought experiment" that has you track how ice is melted by following quanta of heat. It might help to think of the ice in a plastic bag to keep it separate from the water in the container. Equilibrium occurs when the temperatrue of the bag's contents matches the temperature of the surrounding water. A 25 g ice cube is put in 200 g of water at 20 degree C .What will be the final temperature? Reason out the solution instead of calculating it directly! To do this, imagine sending heat from the hot material to the cold material in small amounts un equilibrium is reached. Since ice and water can only coexist in equilibrium at 0 degree C, all of the ice must be melted before the temperature of the cold material can be raised above 0 degree C. At the outset, it is not obvious whether all of the ice will melt. If re remove 80 cal of heat from water, the temperater of water will decrease according to Q mcDT This amounts to a drop of 0.4 C degree if the mass of the water is 200 grams. Adding 80 cal of heat to ice au will melt one gram, leaving 24 grams remaining. Continue until all of the ice has melted or the water has a temperature of zero Celsius. Lf all of the ice melts and the if the water temperature is not 0 degree C, keep transferring heat (possiblly smaller amounts) until the water from the melted ice has the same temperature as the originally 20 degree C water. Keep track of your calculations by making a table like the following Continue the process until you reach equilibrium. What is the final temperature? If it is 0 degree C, how much ice remains?Explanation / Answer
Q = mCdT
80 = 200 x 1 x (20 - T1)
T1 = 19.6
Now we have 1 g of water at 0 deg and 200 g of water at 19.6 deg.
The final temperature, T1f can be found out by using the relation heat lost = heat gained
mC(T1f – 0) = MC(19.6 – T1f)
1 x 1 x T1f = 200 x 19.6 – 200 T1f
T1f = 200/201 x 19.6 = 19.5
80 cal = 201 x 1 x (19.5 – T2)
T2 = 19.104, T2f = 201/202 x 19.104 = 19.01
T3 = 19.01 – 80/202 = 18.614
T3f = 202/203 x 18.614 = 18.52
T4 = 18.52 – 80/203 = 18.128
T4f = 203/204 x 18.128 = 18.04
T5 = 18.04 – 80/204 = 17.65
T5f = 204/205 x 17.65 = 17.56
T6 = 17.56 – 80/205 = 17.17
T6f = 204/205 x 17.17 = 17.09
T7 = 17.09 – 80/206 = 16.69
T7f = 205/206 x 16.69 = 16.62
T8 = 16.62 – 80/207 = 16.231
T8f = 206/207 x 16.231 = 16.15
T9 = 16.15 – 80/208 = 15.77
T9f = 207/208 x 15.77 = 15.69
T10 = 15.697 – 80/209 = 15.31
T10f = 208/209 x 15.31 = 15.23
T11 = 15.23 – 80/210 = 14.85
T11f = 209/210 x 14.85 = 14.78
T12 = 14.78 – 80/211 = 14.4
T12f = 210/211 x 14.4 = 14.34
T13 = 14.34 – 80/212 = 13.96
T13f = 211/212 x 13.96 = 13.89
T14 = 13.89 – 80/213 = 13.52
T14f = 212/213 x 13.52 = 13.45
T15 = 13.45 – 80/214 = 13.08
T15f = 213/214 x 13.08 = 13.02
T16 = 13.02 – 80/215 = 12.65
T16f = 214/215 x 12.65 = 12.59
T17 = 12.59 – 80/216 = 12.22
T17f = 215/216 x 12.22 = 12.16
T18 = 12.16 – 80/217 = 11.79
T18f = 216/217 x 11.79 = 11.74
T19 = 11.74 – 80/218 = 11.372
T19f = 217/218 x 11.372 = 11.32
T20 = 11.32 – 80/219 = 10.95
T20f = 218/219 x 10.95 = 10.90
T21 = 10.90 – 80/220 = 10.54
T21f = 219/220 x 10.54 = 10.49
T22 = 10.49 – 80/221 = 10.13
T22f = 220/221 x 10.13 = 10.08
T23 = 10.08 – 80/222 = 9.72
T23f = 221/222 x 9.72 = 9.68
T24 = 9.68 – 80/223 = 9.32
T24f = 222/223 x 9.32 = 9.28
T25 = 9.28 – 80/224 = 8.92
T25f = 223/224 x 8.92 = 8.88
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