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
Verified Correct Answer: A.
Physical Principles & Numerical Solution
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
Source: Myschool Past Questions Archive (Record #29127)