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

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Homework Help Overview

The discussion revolves around calculating heat transfer in a water bath system involving a cylinder with a piston, where the initial state includes both liquid and solid water at a temperature of 273K. The problem involves determining the heat transfer to the bath as the volume is reduced by a specific amount.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the relationship between volume changes and phase changes of water and ice, questioning the necessary conversions between solid and liquid states to achieve the specified volume reduction.
  • Some participants attempt to apply equations related to pressure, volume, and heat transfer, while expressing uncertainty about their correctness.
  • Questions arise regarding the densities of ice and water at the given temperature and how these relate to the volume change.
  • There is a focus on the signs used in equations, particularly regarding the heat transfer calculations.

Discussion Status

The discussion is ongoing, with participants providing insights and clarifications on the equations and concepts involved. Some guidance has been offered regarding the signs in the equations, but no consensus has been reached on the final approach or calculations.

Contextual Notes

Participants are working within the constraints of the problem statement, which includes specific initial conditions and the requirement to calculate heat transfer without providing complete solutions. There is an emphasis on understanding the relationships between different physical properties and their implications for the problem at hand.

Physgeek64
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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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