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Topic: Total Entropy and Enthalpy of Two Copper Blocks  (Read 7179 times)

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Offline TheToby

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Total Entropy and Enthalpy of Two Copper Blocks
« on: October 29, 2008, 10:31:19 PM »
While studying for a test, I came across the following question.  I realize that heat will flow from the hot copper block into the cold copper block and this will represent a change in heat.  I know that the formula will be q=CΔT with C=0.385 J K-1g-1 .  After this point I get lost in the method.  The question is

Calculate ΔHtotal and ΔStotal when two copper blocks each of mass 10.0 kg, one at 100.0oC, and the other at 0.0oC, are placed in contact in an isolated container.  The specific heat capacity of copper is 0.385 J K-1g-1 and may be assumed constant over the temperature range involved.

Please help me understand this.  Thanks!

Offline Hello12

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Re: Total Entropy and Enthalpy of Two Copper Blocks
« Reply #1 on: February 01, 2009, 12:53:12 PM »
I thought the equation to use would be q=mc change T. If the you assumed the intial temparature of the blocks to room temparature i think the answer may be... q=10000*0.385*75. this will give you a q value. But if you want to work out the enthalpy change you will need to find the moles of the copper and divide this by the q value. I have no idea how to work out the ΔStotal sorry

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