Fork pushed into ice (thermodynamics, clausius clapeyron)

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SUMMARY

The discussion focuses on calculating the temperature decrease required to prevent a garden fork from penetrating a block of ice using the Clausius-Clapeyron Equation. The individual standing on the fork exerts a pressure calculated from their weight (57.8 kg) and the fork's mass (1 kg), distributed across the fork's four prongs with a cross-sectional area of 1 mm² each. The latent heat of fusion of ice (333 × 10³ J kg⁻¹) and the densities of water and ice at 0 °C (1000 kg m⁻³ and 916.7 kg m⁻³, respectively) are critical to determining the melting point under pressure. The assumption of linearity in the phase boundary is acknowledged but deemed necessary for the solution.

PREREQUISITES
  • Understanding of the Clausius-Clapeyron Equation
  • Knowledge of pressure calculations in physics
  • Familiarity with the properties of ice and water, including latent heat and density
  • Basic thermodynamics concepts related to phase changes
NEXT STEPS
  • Explore advanced applications of the Clausius-Clapeyron Equation in phase transitions
  • Study the effects of pressure on the melting point of ice in various conditions
  • Investigate the relationship between temperature, pressure, and the latent heat of fusion
  • Learn about the thermodynamic properties of materials under varying environmental conditions
USEFUL FOR

Students and professionals in physics, particularly those studying thermodynamics, as well as engineers and scientists involved in materials science and phase transition research.

Robsta
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Homework Statement


A person (mass 57.8 kg) tries to penetrate a garden fork (mass 1 kg) into a melting block of ice by standing on it with all his/her weight. Assume that no heat flows from the fork to the ice. The fork has four prongs, and each prong has a square cross section of area 1 mm2 . By how much must the temperature of the ice be lowered to resist penetration? [You may take the density of water and ice at 0 °C to be 1000 kg m−3 and 916.7 kg m−3 , respectively. The latent heat of fusion of ice is 333 × 103 J kg−1 .]

Homework Equations


The Clausius-Clapeyron Equation gives the gradient of the phase boundary between water and ice.

dp/dT = L/TΔV

The Attempt at a Solution


I can work out what the pressure exerted on the ice is, that's fine. I just use the weight of the person and fork and divide it by the area. This pressure will melt the ice into water and penetrate if the ice is not too cold.

I can also calculate L/ΔV from information given at the end of the question. So the gradient of the phase boundary is just (L/ΔV)/T

Perhaps if I say the ice is solid at p = 105Pa and T = 273 K and then assume that the phase boundary is linear, then I can say Δp/ΔT = dp/dT = L/TΔV and solve for T. Then I know the temperature at which the ice is on the point of melting for the given pressure. If the ice is colder than that, it won't be penetrated.

Does that sound right? It seems like I'm making a big assumption about the linearity of the phase boundary over a wide range of pressures.
 
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Robsta said:

The Attempt at a Solution


Does that sound right?
Yes.
It seems like I'm making a big assumption about the linearity of the phase boundary over a wide range of pressures.
How much do you think L changes with temperature and pressure?
How much do you think the density of ice and liquid water change with temperature and pressure?

Chet
 

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