Heat released during freezing of water

[tyneoh]

Homework Statement

A situation is given whereby a thermometer is placed inside acetone of 80C and is taken out. A layer of ice forms around the surface of the thermometer, does the reading of the thermometer change?

Homework Equations

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The Attempt at a Solution

At first I considered during condensation the substance (in this case, water vapour) experiences no change in temperature as heat released is the same as heat formed during the formation of bonds between water molecules. Then I considered that since heat is given out by the water molecules, the thermometer absorbs some of the heat and therefore experiences an increase in its reading. Am I correct to say that the water experiences no change in temperature but the thermometer does?BTW this is not a homework question, it is a question from my examination, and I chose to answer that the reading of the thermometer increase. I have this sinking feeling that I may be wrong

 

[CWatters]

Did you mean 80C or -80C?

Imagine the thermometer was at -100C. When water vapour condenses on the thermometer it will warm the thermometer a bit. However the thermometer will also cool the water and it could cool it below 0C. It would depend on the temperature, mass and material of the thermometer as well as the mass of water.

Imagine what happens if water ice is dropped in liquid nitrogen. The water will warm the N but the N will also cool the water ice.

 

[Chestermiller]

Water vapor will continue to condense and heat the thermometer until the thermometer has reached 0C. After that, the room air will begin to heat the ice from the outside (with the thermometer remaining at 0C) until all the ice has melted. Then, the temperature of the water will begin to rise, and the thermometer temperature will follow.

 

[tyneoh]

Well this question is purely qualitative, there are no calculations involved. Since my description of the question wasn't quite clear I shall copy the question itself

The diagram below shows a thermometer placed in a beaker of acetone and its reading is τ1. When the thermometer is taken out, and icy layer forms around the thermometer and its reading is τ2.Compare τ1 and τ2.

τ1 = τ2
τ1 Greater than τ2
τ1 less than τ2

 

[Borek]

Think in terms of a simple heat balance. An object at -80°C is put in contact with wet air at STP. Water condenses on the surface of the object - heating it up. Water freezes - heating it up. Ice cools down till its temperature is identical with the temperature of the object - object temperature goes up even further.

 

[tyneoh]

Think in terms of a simple heat balance. An object at -80°C is put in contact with wet air at STP. Water condenses on the surface of the object - heating it up. Water freezes - heating it up. Ice cools down till its temperature is identical with the temperature of the object - object temperature goes up even further.

Alright looks like I was correct. Thanks Borek :)