Soldato
- Joined
- 14 Nov 2002
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It wasn't even close, I could tell by immersing my fingers in each cup.
Case (a) will be marginally cooler.
In case (a) you immediately drop the temperature to T=(Th*Mh+Tc*Mc)/(Mh+Mc)
Where Th is temperature of the Hot cup and Mh is the mass of water in the hot cup
Tc and Mc are the same for the cold cup. (temperatures need to be in K)
From this point onwards your rate of heat transfer will be increased very slightly due to the increased surface area that the heat can escape from the cup.
In case (b) you don't get that initial drop in temperature and the benefit of slightly increased surface area. Yes, you will have a higher initial rate of heat loss as the temperature will be warmer but that rate will decrease as you drop the temperature. Q=k*Area*(Th-Tair)
I'm also assuming that the density of the water does not change significantly with temperature.
from your post, are you just guessing a)? Because both of your versions of events are purely qualitative, so how can you make a quantitive distinction?
The important thing is the temperature difference, as heat transfer is proportional to it. Higher the temp difference, faster the heat transfer. Hotter water will transfer heat to the room faster, so by not adding the cooler water to the beaker of hot water it will lose more heat to the room by itself. Add the cool water in at the last minute. Option b).
Because i'm using sound thermodynamic reasoning. We were only asking for a qualitative answer. I could go away and actually calculate the answer but I would have to make some assumptions based on what the thermal constants would be. Even if i was miles out on the thermal constants the equations will still tell me that case (a) will end up cooler.