A calorimeter of thermal capacity 80j contains 20g of wat...
A calorimeter of thermal capacity 80j contains 20g of water at 25°C. Water at 100°C is added so that the final temperature of the set-ups is 50°C. The amount of water added is (Heat capacity of water = 4.18J/g/°C)
20g
25g
45g
50g
100g
Correct answer is A
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