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200 g of hot water at 80^@C is added to ...

200 g of hot water at `80^@`C is added to 300 g of cold water at `10^@`C. Neglecting the heat taken by the container, calculate the final temperature of the mixture of water. Specific heat capacity of water = 4200 J `kg^(-1)K^(-1)`

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To solve the problem of finding the final temperature of a mixture of hot and cold water, we can use the principle of calorimetry, which states that the heat lost by the hot water will be equal to the heat gained by the cold water. ### Step-by-Step Solution: 1. **Identify the given data:** - Mass of hot water (m₁) = 200 g = 0.2 kg (since 1 g = 0.001 kg) - Initial temperature of hot water (T₁) = 80 °C - Mass of cold water (m₂) = 300 g = 0.3 kg ...
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