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?

## Determining calorimeter constant

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- Dhamnekar Winod
- Sr. Member
**Posts:**35**Joined:**Sat Nov 21, 2020 10:14 am**Location:**Mumbai[Bombay],Maharashtra State,India

### Determining calorimeter constant

Last edited by Dhamnekar Winod on Wed Jan 27, 2021 9:03 pm, edited 1 time in total.

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- ChenBeier
- Distinguished Member
**Posts:**537**Joined:**Wed Sep 27, 2017 7:25 am**Location:**Berlin, Germany

### Re: Determining calorimeter constant

The video is correct.

The generell calculation is

cCal = (m2* cH

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.

The generell calculation is

cCal = (m2* cH

_{2}O*( T_{1}- Tmix)-m1*cH_{2}O*(Tmix-T_{1}))/(Tmix-T_{1})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.