Computing calorimeter constant

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Dhamnekar Winod
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Computing calorimeter constant

Post by Dhamnekar Winod »

Calculate the calorimeter constant if 25.0 g of water at 60.0 °C was added to 25.0 g of water at 25.0 °C with a resulting temperature of 35.0 °C?

Solution:

If the constant were zero, the final temperature of the water would be 42.5 °C. So the amount of heat used by the calorimeter to heat from 25 to 35 is:

(25.0) (4.184) (7.5) = 784.5 J.
Since the constant is Joules/degree, the constant is

784.5 J / 10.0 °C = 78.4 J/°C (to three sig figs).

My question:- I don't understand what is the meaning of underlined text? Would any member of chemistry forum of WebQc.org answer this question?
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ChenBeier
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Re: Computing calorimeter constant

Post by ChenBeier »

If you mix the same amount of water with different temperatures then the mixing temperature would be the arithmetical middle of it .inyour case (60 + 25) /2 = 42,5 °C.
But the calorimeter has a constant. The achieved temperature was 35 "C. So it means 7,5 K less as 42,5 °C.
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ChenBeier
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Re: Computing calorimeter constant

Post by ChenBeier »

Yo have to take the total mass of water.
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