Answer : - 59.5 °C
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Given :-
1. 63.4 g water at temperature 80.3 °C
2. 39.5 g water at temperature 26.1 °C
The specific heat of water = C = 4.184 J/g. °C
Formula :-
or
i.e. or
...(1)
where,
q = heat gain , - q = heat lost
m = mass
C = specific heat
Tf = final temperature
Ti = initial temperature
When two water sample mix with each other
i) sample with high temperature (80.3 °C) lost heat and its temperature decrease
ii) sample with low temperature (26.1 °C) gain heat and its temperature increase
And we have, ...(2)
First using equation (1) we can calculate, and
Thus,
i.e.
i.e.
And
Therefore from equation (2) we have
i.e.
i.e.
i.e.
i.e.
i.e
i.e.
i.e.
i.e.
i.e.
(up to three significant figure)
Therefore, the final temperature when 63.4 g of water at 80.3 °C is mixed with 39.5 g of water at 26.1 °C is 59.5 °C
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What is the final temperature when 63.4 g of water at 80.3°C is mixed with 39.5...
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