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Question:

A metal ball of mass 0.1kg is heated up to 500oC and dropped into a vessel of heat capacity 800 JK⁻¹ and containing 0.5kg water. The initial temperature of water and vessel is 30oC. What is the approximate percentage increment in the temperature of the water?

30

25

15

20

Solution:

0.1 × 400 × (500 − T) = 0.5 × 4200 × (T − 30) + 800(T − 30) → 40(500 − T) = (T − 30)(2100 + 800) → 20000 − 40T = (T − 30)(2900) → 20000 − 40T = 2900T − 87000 → 20000 + 87000 = 2940T → 107000 = 2940T → T = 36.4oC ΔT = 36.4 − 30 = 6.4 ΔT/30 × 100 = 6.4/30 × 100 = 20