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A metal ball of mass 0.1 kg is heated upto 500°C and dropped into a vessel of heat capacity 800 J K⁻¹ containing 0.5 kg water. The initial temperature of water and vessel is 30°C. What is the approximate percentage increment in the temperature of the water? [Specific Heat Capacities of water and metal are, respectively, 4200 J kg⁻¹ K⁻¹ and 400 J kg⁻¹ K⁻¹]

1. 20%
2. 25%
3. 15%
4. 30%

Correct answer: 20%

Solution

0.1*400*(500-Tf) = (0.5*4200 + 800)*(Tf-30). This gives Tf ~ 36.4 C, so the water temperature rises by ~6.4 C. Percentage increment = 6.4/30 * 100 ~ 21%, i.e. about 20%.

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