What is the Total Heat Capacity of the Calorimeter?

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SUMMARY

The discussion focuses on calculating the total heat capacity of a calorimeter when mixing two water samples at different temperatures. The initial calculations provided by the user incorrectly summed the heat exchanges instead of applying the principle of conservation of energy. The correct approach involves using the equation ΔQ(hot) = ΔQ(calorimeter) + ΔQ(cold) to account for heat loss from the hot water to both the calorimeter and the cold water. The final heat capacity of the calorimeter should be calculated as 493.24 J/K, correcting the user's misapplication of the heat transfer equations.

PREREQUISITES
  • Understanding of heat transfer principles
  • Familiarity with calorimetry concepts
  • Knowledge of specific heat capacity calculations
  • Basic algebra for solving equations
NEXT STEPS
  • Review the principles of conservation of energy in calorimetry
  • Learn about heat capacity and its calculation methods
  • Study the effects of temperature on specific heat capacity
  • Practice solving calorimetry problems with varying conditions
USEFUL FOR

This discussion is beneficial for students studying thermodynamics, chemistry enthusiasts, and anyone involved in calorimetry experiments or heat transfer calculations.

gogeta2006
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Homework Statement



50ml of water at 49.6 C were mixed with 50ml of water at 25.1 C in a calorimeter also at 25.1 C. The final temperature was 30.1 C Assuming that neither the density of water nor its specific heat capacity change with temperature, calculate the total heat capacity of the calorimeter.


Homework Equations



Density of water = 1.00 g/mL
Specific heat capacity = 4.18 J / g * K

The Attempt at a Solution



q (heat given up by water) = 50ml * (49.6-30.1)
= 840 cal

q (heat absorbed by cold water) = 50ml (30.1-25.1)
= 250 cal

Heat absorbed by calorimeter = 250 + 840 = 1090 cal

Ccal = qcal / delta T
= 590 / (30.1-25.1)
= 118 K

The answer is supposed to be 493.24 J/K ... but i am not getting that.
Please someone please show me how to correct this.

Thank you.
 
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Heat absorbed by calorimeter = 250 + 840 = 1090 cal
Woah!

Heat is lost from the hot water to the calorimeter AND the cold water. One cannot at the heat of the hot and cold water.

Try ΔQ(hot) = ΔQ(calorimeter) + ΔQ(cold)
 
Astronuc said:
Woah!

Heat is lost from the hot water to the calorimeter AND the cold water. One cannot at the heat of the hot and cold water.

Try ΔQ(hot) = ΔQ(calorimeter) + ΔQ(cold)

I tried doing that and the answer is 118 (which is still INCORRECT)...i wrote plus there where it should be minus.
 
Last edited:
Show your detailed work then.
 

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