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

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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

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 =

Oct 10th, 2015

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

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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Oct 10th, 2015
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Oct 10th, 2015
Nov 18th, 2017
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