Heat loss

bit of a bogey

This is a good little brain teaser for ye physicists! I'm looking for a bit of help on how to calculate the rate of which heat is lost.

If I have a beaker filled with water that has been heated to a temperaure of just say 60C. The heated water is then left exposed to room temperature. How can I calculate the temperature after 300 seconds? I dont know how much energy was used to heat the water to 60C in the first place. Do I have to consider the specific heat capacity of my beaker or would it be the thermal conductivity? Where would this come into my calculations? I know Q=mc(dT) but i dont really know what to do with it in this situation.

Any ideas? Thanks!

bit of a bogey

NO?

Professor_Fink

bit of a bogey wrote: »

NO?

Actually, it is somewhat more complicated than that. Heat loss is invariably a function of surface area, rather than volume, so you need to calculate the surface area of the water. Usually the loss is directly proportional to the surface area, but will depend on the temperature of the surrounding environment, and the thermal conductivity of the material at the interface with the water.

You can make some crude approximations to eliminate some of these variables, but for an accurate calculation quite a bit of detail is needed (even the volume of the room would play a very small role, since the cooling liquid heats the room).

Capt'n Midnight

worse you need also to calculate heat loss due to evaporation which depends on the area at the top , velocity of drafts , saturation of the air.

how long is a piece of string ?

to make it simplier you could make some assumptions about the environment , rule out convection , assume all the heat loss is through the bottom of the vessel on to an surface

bit of a bogey

yeh im kinda realizing its not as straight forward. Any ideas of where to even start?

c-note

your heat loss will depend on the temprature gradient.
so you need to define what "room temprature" is

as you loose heat your temprature gradient changes,
(i can feel some calculus coming on)

you may want to investigate lumped capacitance?

Professor_Fink

c-note wrote: »

your heat loss will depend on the temprature gradient.
so you need to define what "room temprature" is

as you loose heat your temprature gradient changes,
(i can feel some calculus coming on)

you may want to investigate lumped capacitance?

There is a heat loss law that goes something like dE/dt = c(T1-T2), where T1 and T2 are the environment and sample temperatures respectively. Unfortunately calculating c is incredibly hard (depending on all the factors we mentioned earlier in the thread.

Capt'n Midnight

http://www.efunda.com/Materials/water/steamtable_sat.cfm pop in 60 degrees to the properties of water / evaporated water vapour. also pop in 21 degrees to get a final result or whatever you decide to use as room temperature

Professor_Fink

Capt'n Midnight wrote: »

http://www.efunda.com/Materials/water/steamtable_sat.cfm pop in 60 degrees to the properties of water / evaporated water vapour. also pop in 21 degrees to get a final result or whatever you decide to use as room temperature

That's not going to give you a particularly relevant number. It's just a tiny part of the overall calculation.