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specific heat capacity

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Joel:
Hello! I'm new here as you can see, and I have a question :)

A 150.0g piece of a metal at 75.0 degrees C is added to 150.0g of water at 15.0 degrees C. The temperature of the water rises to 18,3 degrees C.

How can I now calculate the specific heat capacity of the metal, assuming that all the heat lost by the metal is gained by the water? I've tried for a while now, and I can't get it right.

sdekivit:

--- Quote from: Joel on October 05, 2005, 02:34:49 PM ---Hello! I'm new here as you can see, and I have a question :)

A 150.0g piece of a metal at 75.0 degrees C is added to 150.0g of water at 15.0 degrees C. The temperature of the water rises to 18,3 degrees C.

How can I now calculate the specific heat capacity of the metal, assuming that all the heat lost by the metal is gained by the water? I've tried for a while now, and I can't get it right.

--- End quote ---

the heat released by the reaction is taken up by the water . Thus

Q(metal) = Q(water)

-->

c * m * delta T = c * m * delta T

Donaldson Tan:
assuming 100% heat transfer,

heat gain by water = heat loss by metal

heat gain by water = mass of water X heat capacity of water X temperature increment of water.

heat loss by metal = mass of metal X heat capacity of metal X temperature decrement of metal

the temperature of the metal after cooled down will be the same as the final temperature of water because only at thermal equilibrium then there will be no heat transfer.

Joel:
Thanks to both of you! It all makes sense to me now and I got the problem solved :)

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