How Do You Calculate the Specific Heat Capacity of an Object?

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SUMMARY

The discussion focuses on calculating the specific heat capacity of an object when it is placed in water, using the principle of conservation of energy. The specific heat capacity of water is given as 4200 J/kg·K. The problem involves a 0.6 kg object at 6°C placed in 1.2 kg of water at 28°C, resulting in a final temperature of 26°C. The calculations involve setting the heat lost by the water equal to the heat gained by the object, leading to the equations Q1 = m_water * c_water * ΔT_water and Q2 = m_object * c_object * ΔT_object.

PREREQUISITES
  • Understanding of specific heat capacity and its formula
  • Basic knowledge of thermal energy transfer
  • Ability to perform calculations involving mass, temperature change, and energy
  • Familiarity with the concept of conservation of energy in thermodynamics
NEXT STEPS
  • Calculate the specific heat capacity of the object using the derived equations
  • Explore the concept of thermal equilibrium in heat transfer
  • Learn about the implications of heat loss in real-world scenarios
  • Investigate different methods for measuring specific heat capacity experimentally
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Students studying thermodynamics, physics enthusiasts, and anyone involved in calorimetry or heat transfer calculations.

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The question I'm currently working on:
When 0.6 kg of an object at 6 degree Celsius is put inside a beaker of water of mass 1.2Kg at 28 degree Celsius, the final temperature of water is 26 degree Celsius. Assuming no heat lost to the surroundings and the heat absorbed by the beaker is negligible, calculate the specific heat capacity of the object. (Specific heat capacity of water is 4200 J KG-1 K-1)

So far.. MY solution is:

Thermal energy, Q= specific heat capacity of water x mass of object and water x the change in temperature
which is: 4200 x 1.8 x (-8) = -60480

I am really clueless right now.. Hope you guys are able to help and guide me with this one..
 
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Nearly correct,
The change in energy of the water is:
Q1 = specific heat of water * mass of WATER * change in temperature of water

Then you can use this with:
Q2 = specific heat of object * mass of OBJECT * change in temperature of object

Since no heat is lost, Q1 lost by water equals the Q2 gained by object.
 

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