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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His Name is Tiamiyu Abdulbazeet Olawale. He is a student at the University of Ilorin studying Electrical and Electronics Engineering. He likes to give the latest updates about Nigeria and international institutions as he aims at helping people to lay their hands on the right information.