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)
m = 20g