How can I find the heat capacity of Al, i think I am using the wring equations.

Chemistry
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A student has calibrated his/her calorimeter and finds the heat capacity to be 15.4 J/°C. S/he then determines the molar heat capacity of aluminum. The data are: 24.0 g Al at 100.0°C are put into the calorimeter, which contains 99.1 g H2O at 19.8°C. The final temperature comes to 23.6°C. Calculate the heat capacity of Al in J/mol·°C.

Oct 10th, 2015

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Solution

 Since the heat capacity has units of J/C find the Joules 15.4J/C x (23.6C-19.8C) 

15.4x 3.8  that gives you Joules 58.52J  gained by calorimeter

(3.8 J/g·°C) x (99.1 g) x (23.6 - 19.8)°C = 1431.004 J gained by the water

58.52 J + 1431.00 J = 1489.52 J total lost by the Al


you want the heat capacity of Al J/g C 
Take the J from before and divide it by the moles of Al (25.6g divide by the molar mass) and the temp change of the Al (100C-23.6C) 

Moles of Al = 24/26.98

(1489.52 J) / (24.0 g Al / (26.98154 g Al/mol)) / (100.0 - 23.6)°C =


http://www.kentchemistry.com/links/Energ...


Please let me know if you need any clarification. I'm always happy to answer your questions.
Oct 10th, 2015

The answer is 21.92 joules 

Oct 10th, 2015

Where did the 3.8 J/g°C come from ?  I thought the 3.8 was a temperature difference.
Also, it is saying the answer is wrong...

Oct 10th, 2015

I am pasting the  revised answer giev me one second '


Oct 10th, 2015

REVISED ANSWER 

 Since the heat capacity has units of J/C find the Joules 15.4J/C x (23.6C-19.8C) 

15.4x 3.8 that gives you Joules 58.52J  gained by calorimeter

(4.184J/g·°C) x (99.1 g) x (23.6 - 19.8)°C = 1575.61J gained by the water (4.184 specific heat of water)

58.52 J + 1575.61 J = 1634.13 J total lost by the Al


Oct 10th, 2015


you want the heat capacity of Al J/g C 
Take the J from before and divide it by the moles of Al (25.6g divide by the molar mass) and the temp change of the Al (100C-23.6C) 

Moles of Al = 24/26.98

(1634.13J) / (24.0 g Al / (26.98154 g Al/mol)) / (100.0 - 23.6)°C = 24.04 joules/mole/degreeC


Oct 10th, 2015

so 24 J 

Oct 10th, 2015

yes if you round it up 

Oct 10th, 2015

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