Melting Ice: 200g Steel at 150°C on 0°C Ice

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SUMMARY

The discussion focuses on the thermal interaction between 200g of steel at 150°C and a block of ice at 0°C. The calculation for the amount of ice melted is determined using the formula Q=mcΔT, resulting in approximately 0.041317 kg of ice melted. Additionally, the participant calculates the depth the steel would sink into the ice, concluding it would be approximately 0.89 cm based on the volume of the melted ice and the area of the circular steel piece.

PREREQUISITES
  • Understanding of thermodynamics principles, specifically heat transfer.
  • Familiarity with the specific heat capacity of materials, particularly steel and ice.
  • Knowledge of basic geometry for calculating areas and volumes.
  • Proficiency in using the mass-energy equivalence formula Q=mcΔT.
NEXT STEPS
  • Study the specific heat capacity values for various materials, including steel and ice.
  • Learn about the phase change of water and the latent heat of fusion.
  • Explore advanced thermodynamic calculations involving multiple materials.
  • Investigate the effects of different temperatures on phase changes in substances.
USEFUL FOR

Students in physics, engineers working with thermal systems, and anyone interested in thermodynamics and heat transfer calculations.

Ry122
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200g of steel was placed on a big block of ice.
The ice was 0 degrees C.
The metal was 150 degrees C.
How much ice gets melted?

My attempt:
Q=mc(changeinT)
Q=.2(460)150
Q=13800

Q=ML
13800=M(3.34x10^5)
M=.041317 kg is melted
Is this correct?
 
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Looks fine to me.
 
If this piece of metal was circular with a radius of 4 cm. How deep would it sink into the ice?

My attempt:
M=.041317 kg is melted

Area of circle = pie x r^2 = pie x .04^2 = 5.02 x 10^-3 m
Now setup an equation to find the height of the cylinder made.
Find what volume this would be.
v=m/d
= .041317/920
= 4.49 x 10^-5
5.02 x 10^-3 x height = 4.49 x 10^-5
height = 8.944 x 10^-3m
= .89cm
Is this correct?
 

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