Specific Heat of Ice and Copper

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SUMMARY

The discussion centers on calculating the mass of copper required to melt a 200-gram block of ice at -10.0°C and raise the temperature of the resulting water to 15.0°C. The initial calculation incorrectly assumed a single-step process, while the correct approach involves three distinct steps: heating the ice to its melting point, melting the ice, and then heating the water. The specific heats used are 2.1 KJ/kg*K for ice, 4.18 KJ/kg*K for water, and 0.385 KJ/kg*K for copper, along with the heat of fusion of water.

PREREQUISITES
  • Understanding of specific heat capacity
  • Knowledge of heat transfer principles
  • Familiarity with the heat of fusion of water
  • Ability to perform unit conversions (grams to kilograms)
NEXT STEPS
  • Study the specific heat capacity of water (4.18 KJ/kg*K)
  • Learn about the heat of fusion of water (334 KJ/kg)
  • Explore thermodynamic principles related to heat transfer
  • Practice multi-step calorimetry problems
USEFUL FOR

Students studying thermodynamics, physics educators, and anyone interested in calorimetry and heat transfer calculations.

liz_p88
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Homework Statement



A student wants to melt a 200 gram block of ice that is at minus 10.0°C and turn it into liquid water at 15.0°C. The student plans to do this by heating copper pellets to 100.0°C and then dropping them onto the block of ice in a thermally insulated container. How many grams of copper will the student need to use?

Homework Equations



mcΔT = - [mcΔT]

The Attempt at a Solution



(200g)(2.1 KJ/kg*K)(25°C) = -(m)(0.385 KJ/kg*K)(-85°C)

m = 320.86 g

DID I DO THIS CORRECTLY?
 
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No. There are three steps.
1)heat ice from minus 10.0°C to its melting point
2)melt ice into water
3)heat water to 15.0°C
you will need the specific heats of ice, water, and copper and the heat of fusion of water.
 

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