Thermodynamics: Finding Specific Heat of Ice

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SUMMARY

The discussion focuses on calculating the specific heat of ice using the formula Q = mcΔT. Given a mass of 0.25 kg of ice, an energy supply of 530 W over 15 seconds, and a temperature change of 15 degrees Celsius, the calculated specific heat (c) is determined to be 2120 J/(kg·°C). The method employed is correct, although it is suggested to include units for clarity.

PREREQUISITES
  • Understanding of thermodynamic principles
  • Familiarity with the specific heat capacity formula Q = mcΔT
  • Basic knowledge of units of energy (Joules) and power (Watts)
  • Ability to perform unit conversions and algebraic manipulations
NEXT STEPS
  • Study the concept of specific heat capacity in different materials
  • Learn about the implications of energy transfer in thermodynamics
  • Explore the relationship between power, energy, and time in thermal systems
  • Investigate real-world applications of specific heat in climate science and engineering
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Students studying thermodynamics, physics educators, and anyone interested in understanding heat transfer and specific heat calculations.

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Homework Statement


It takes 15 seconds to warm ice 15 degrees ( -15 degrees celsius to 0 degrees celsius). The ice has a mass of .25 kg. The energy is supplied at 530 W to the ice.

Homework Equations


Q = mc delta T


The Attempt at a Solution


I plugged in all the information for Q
(15 seconds * 530 W) = .25 * c * 15 degrees
which ends up with c = 2120

Did I do this correctly
 
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You might want to add some units - but correct method
 

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