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Standard X

Physics

Specific Heat Capacity

200 g of hot ...

Question

200 g of hot water at $80^{0}$ C is added to 300 g of cold water at $10^{0}$ C. Neglecting the heat taken by the container, calculate the final temperature of the mixture of water . Specific heat capacity of water = 4200 J $k g^{- 1} K^{- 1}$

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Solution

Given,
Specific heat of water $S_{W} = 4.2 k J / k g K$
Mass of cold, water, $M_{C} = 0.3 k g$
Mass of hot, water, $M_{H} = 0.2 k g$
If final temperature is $T_{f}$
Heat Loos = heat gain
$M_{H} S_{W} (80 - T_{f}) = M_{C} S_{W} (T_{f} - 10)$
$T_{f} = \frac{M_{H} S_{W} \times 80 + M_{C} S_{W} \times 10}{M_{H} S_{W} + M_{C} S_{W}}$
$T_{f} = \frac{0.2 \times 4.2 \times 80 + 0.3 \times 4.2 \times 10}{0.2 \times 4.2 + 0.3 \times 4.2}$
$T_{f} = 38^{o} C$
Final temperature of mixture is $38^{o} C$

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200 g of hot water at 800C is added to 300 g of cold water at 100C. Neglecting the heat taken by the container, calculate the final temperature of the mixture of water . Specific heat capacity of water = 4200 J kg−1K−1

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200 g of hot water at $80^{0}$ C is added to 300 g of cold water at $10^{0}$ C. Neglecting the heat taken by the container, calculate the final temperature of the mixture of water . Specific heat capacity of water = 4200 J $k g^{- 1} K^{- 1}$