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Question
Numerical Problem.
What could be the final temperature of a mixture of 100 g of water at 90 °C and 600g of water at 20°C.
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Solution
To find final temperature: ∆Q = mc
lOOg of water originally at 90°C will loose an amount of heat,
∆Q = mc ∆T
∆Q = 100 × c × (90 – T)
The same amount of heat will be absorbed by 600g of water originally at 20°C to raise its temperature to T.
∆Q = 600 × c × (T – 30)
600C (T – 20°) = 100C (90° – T)
6T – 120° = 90° – T
6T + T = 120° + 90°
7T = 210° ⇒ T = 210/7
T = 30°C
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