Determining calorimeter constant

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Dhamnekar Winod
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Determining calorimeter constant

Post by Dhamnekar Winod »

https://www.chemteam.info/Thermochem/Vi ... ant-02.mp4

My answer to the problem in the above video:

If the constant were zero, the final temperature of the water would be 45.0 °C. So the amount of heat used by the calorimeter to heat from 21.3 to 43.3 is:

(50.0) (4.184) (1.7) = 355.6 J.
Since the constant is Joules/degree, the constant is

355.6 J / 22.0 °C = 16.2 J/°C (to three sig figs).

I want to know which answer is correct? The answer given in the video or my answer with aforesaid working?
Last edited by Dhamnekar Winod on Wed Jan 27, 2021 9:03 pm, edited 1 time in total.
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ChenBeier
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Re: Determining calorimeter constant

Post by ChenBeier »

The video is correct.

The generell calculation is

cCal = (m2* cH2O*( T1- Tmix)-m1*cH2O*(Tmix-T1))/(Tmix-T1)

cCal = (50 g*4,184 J/g°C*(68,7-43,3)°C-50g*4,184J/g°C*(43,3-21,3)°C/(43,3-21,3)°C
cCal = (209,2*25,4 -209,2*22) /22 = 209,2* 3,4/22 = 32,3 J

The calculation in your first exercise was wrong, because you have to consider the total mass of water which is 2 times 25 g = 50 g.
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