Question : Thermal Equilibrium in 3 substances

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SUMMARY

This discussion focuses on calculating the thermal equilibrium temperature when three beakers of water at different temperatures are mixed. Beaker A contains 0.5 kg of water at 30°C, Beaker B has 0.3 kg at 40°C, and Beaker C holds 0.2 kg at 20°C. The heat capacity of water is given as 4.2 joules/gram°C. The final temperature (T) can be determined by setting the sum of the heat changes for each beaker equal to zero, leading to the equation: 0.5 × 4200 × (T - 30) + 0.3 × 4200 × (T - 40) + 0.2 × 4200 × (T - 20) = 0.

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Note : My syllabus only go until thermal equilibrium in 2 objects , i am wondering what will happen and how to calculate if there are 3 substances in contact. So i formed a question on my own hoping it can be solved.

*my teach refused to show me calculations saying it is unrelevant to exam.

given 3 beakers of water at room temperature and pressure are to mixed into one beaker
each beaker labelled A : has 0.5kg mass at 30°C
B : has 0.3kg mass at 40°C
C : has 0.2kg mass at 20°C

let heat capacity of water = 4.2 joule/gram °C

assume heat transfer only occurs among the water with no heat loss to surroundings.
what is the temperature of the water when it reaches thermal equilibrium?

if there are other examples of thermal equilibrium in 3 substances with shown calculations from other websites , mind to post here one? i searched but found nothing , therefore i post a question here.
 
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The ideas and concepts are the same as for two fluids; i think where you may be having trouble is in the mathematics and formulation of the problem?

Start by defining T to be the final temperature of the mixture. You can then write expressions for the change in energy of each mass of water. For example, the 0.5kg mass has the following expression:

\Delta H_A=0.5\times 4200\times (T-30)

You can do this for each mass of water. Then have a think about how each of these energy changes should be related.
 

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