Calculating Heat Transfer in a Water Bath with a Changing Volume and Temperature

AI Thread Summary
The discussion centers on calculating heat transfer in a water bath involving a cylinder with a piston, initially containing 10kg of water and ice at 273K. Participants explore the relationship between pressure, volume, and temperature, using equations related to heat transfer and phase changes. Key questions include the densities of ice and water at this temperature and the volume change resulting from converting ice to water. The calculation suggests that approximately 9.58 grams of ice must be converted to achieve the required volume reduction, leading to a heat transfer of about 3191.25 J. The importance of sign conventions in these calculations is emphasized throughout the discussion.
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Homework Statement


A cylinder is fitted with a piston and is in thermal contact with a heat bath at 273K. Initially the volume in the cylinder is filled with 10kg of pure H2O and about half of this is liquid and the other half is solid. The piston is lowered so as to reduce the volume by 2 × 10−5 m3. What is the sign and magnitude of the heat transfer to the bath?

Homework Equations

The Attempt at a Solution


##\frac{dp}{dT}=\frac{L}{T\Delta V}##
##p_2-p_1= \frac{L}{\Delta V} ln{\frac{T_2}{T_1}}##
but ##T_1=T_2## hence ##p_1=p_2=constant##
##dQ=dU+pdV=C_VdT+pdV=pdV##
##\Delta Q=p\Delta V##

I feel this is probably all wrong, but i can't see how else to do it- any help would be very much appreciated.
 
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What are the densities of ice and water at 273K? How much ice must be converted to water, or vice versa, to reduce the total volume by 2 x 10-5 m3?
 
mjc123 said:
What are the densities of ice and water at 273K? How much ice must be converted to water, or vice versa, to reduce the total volume by 2 x 10-5 m3?

##V=\frac{M}{\rho}##
##\Delta V= \delta \big{(} \frac{1}{\rho_l} + \frac{1}{\rho_s} \big{)}##
where ##\delta## is the mass that is converted from solid to liquid.
## \delta= 9.58 \times 10^{-3}##

Can we then use ##lM=T\Delta S = \Delta Q##
##\Delta Q = l \delta = 3191.25 J## ?

where l is the specific latent heat of fusion of ice
 
Last edited:
Physgeek64 said:
ΔV=δ(1ρl+1ρs)ΔV=δ(1ρl+1ρs)\Delta V= \delta \big{(} \frac{1}{\rho_l} + \frac{1}{\rho_s} \big{)}
(Why does it always look wrong in quotes?)
Why the plus sign? You want the difference in volume between liquid and solid.
 
mjc123 said:
(Why does it always look wrong in quotes?)
Why the plus sign? You want the difference in volume between liquid and solid.
Ahh yes! So with the minus sign, does this look right?

Thanks
 
mjc123 said:
(Why does it always look wrong in quotes?)
Selecting text to quote appears to translate away from the underlying LaTeX. To get it to display properly, you have to use the Reply "button" on the lower right-hand side of the post. That will quote the entire post, but you can trim it down if you just want to quote a part.
 
Physgeek64 said:
Ahh yes! So with the minus sign, does this look right?
Should be. Be careful with the sign of Q.
 
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