Physics

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.

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asked by Maria
  1. 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

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    posted by drwls

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