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)
Correct Answer: Option A
Heat lost = heat gained
mcθ = cθ + mcθ
m x 4.18 x (100 – 50) = 80(50 -25) + [20 x 4.18] x (50 – 25)
Do the math;
m = 20g
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