2008/10/17

Coupled systems and heat of reaction

Today we worked through a coupled systems problem in which a warm block of lead transferred energy to a cooler block of silver. In working through the algebra, I made an error somewhere and came up with the wrong answer, so let's try to correct it here. The set-up was OK, so we got to the following equation:
(0.128 J/g.degC)(17.89g)(Tfinal-42.73degC) = -(0.235 J/g.degC)(23.13g)(Tfinal-15.42degC)
This is set up correctly, now we have to work through the algebra:
(Tfinal-42.73degC) = -{(0.235 J/g.degC)(23.13g)}/{(0.128 J/g.degC)(17.89g)}(Tfinal-15.42degC)
Turning the great big ugly fraction into a number, we get:
(Tfinal-42.73degC) = -(2.3737)(Tfinal-15.42degC)
Distributing the -2.3737:
(Tfinal-42.73degC) = -(2.3737Tfinal)+(2.3737)(15.42degC)
Moving the Tfinal terms to one side and the numerical terms to the other:
Tfinal + 2.3737Tfinal = (2.3737)(15.42degC) + 42.73degC
Factoring out Tfinal and solving:
Tfinal = {(2.3737)(15.42degC) + 42.73degC}/{1 + 2.3737} = 23.5degC

There is a new Mastering Chemistry posted, due Tuesday. Have a great weekend.

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