The final temperature of the 200 g of water will be the same as that of the 100g of hot water that was added. That is when thermal equilibrium exists.
At the final temperature T, the heat gained by the initially cold water will equal the heat lost by the hot water.
C*200g*(T-20) = C*100g*(80-T)
The specific heat C cancels out.
Solve for T, the final equilibrium temperature
2(T-20) = 80-T
3T = 120
T = 40 C
If 200g of water at 20˚ C is combined with 100g of water at 80˚ C, what is the final temperature of the 200g of water?
How do you do this??? steps by steps??
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