a calorimeter of thermal capacity 80j contains 20g of water at 25oC. Water at 100oC is added so that the final temperature of the set-ups is 50oC. The amount of water added is (Heat capacity of water = 4.18J/g/oC)
A.
20g
B.
25g
C.
45g
D.
50g
E.
100g
Correct Answer: Option A
Explanation
Heat lost = heat gained
mc\(\theta\) = c\(\theta\) + mc\(\theta\)
m x 4.18 x (100 - 50) = 80(50 -25) + [20 x 4.18] x (50 - 25)
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