A 669 g sample of water at 83 degrees celsius is mixed with 486 g of water at 23 degrees celsius. What is the final temperature of the mixture? Assume no heat loss to the surroundings.
heat lost by the hot water = heat gained by the cold water
C*669*(83 - T) = C*486*(T - 23)
The specific heat C cancels out.
83-T = 0.726*(T-23)
Solve for equilibrium T