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Author Topic: URGENT Calorimetry concept clarification  (Read 4129 times)

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URGENT Calorimetry concept clarification
« on: October 28, 2007, 04:53:31 PM »

so, I was doing this problem to study for tomorrow's test...

"When a 6.50-g sample of solid Sodium Hydroxide dissolves in 100.0g of water in a coffee cup calorimeter, the temperature rises from 21.6C to 37.8C. Calculate delta H in Kg/mol NaOH for the solution process

NaOH(s) ->Na+(aq) + OH-(aq)

Assume that the specific heat of the solution is the same as that of pure water"

This is what I did, I multipled 106.5g x 4.184J/gC x 16.2C to get the heat or delta H, which is 7.2187KJ. I then divided that by the number of moles of NaOH, which gives me the answer of 44.4KJ/mol NaOH...

However, the answer is -44.4KJ/mol NaOH... What in my calculations did do wrong? Delta T is positive because the final temperature is larger than the initial temperature, so I don't know what I did wrong...


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Re: URGENT Calorimetry concept clarification
« Reply #1 on: October 28, 2007, 05:14:18 PM »

The reaction is exothermic so delta H should be negative. 

The equation Q = m x C x delta T is used correctly here but Q (enthalpy) stands for heat added.  In this situation heat is lost (exothermic) to the surroundings.  This equation assumes you understand which way heat is transferred.

In the realm of scientific observation, luck is only granted to those who are prepared. -Louis Pasteur


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Re: URGENT Calorimetry concept clarification
« Reply #2 on: October 31, 2007, 01:40:01 PM »

just look at the temperature, if the final temp is higher than initial, you know the energy was transferred to the water( so the system, the reaction was exothermic) so you put negative sign for that because it gave away heat.
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