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Method of Mixtures

300 g of coco...

Question

$300 g$ of coconut oil at $50^{\circ} C$ is mixed with $500 g$ kerosene at $10^{\circ} C$ . Determine the temperature of the mixture. (Specific heat of coconut oil = $2100 J k g^{- 1} K^{- 1}$ and specific heat of kerosene = $2000 J k g^{- 1} K^{- 1}$ .

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Solution

Given:
Mass of coconut oil, $m_{C} = 300 g = 0.3 k g$ .
Mass of kerosene, $m_{K} = 500 g = 0.5 k g$ .
Specific heat of coconut oil, $S_{C} = 2100 J / k g K$ .
Specific heat of kerosene , $S_{K} = 2000 J / k g K$
Temperature of coconut oil, $T_{C} = 50^{\circ} C$ and
Temperature of kerosene, $T_{K} = 10^{\circ} C$ .
Let the temperature of the final mixture be $T_{m}$
We know that, heat lost by coconut oil = heat gained by kerosene
$m_{C} S_{C} (T_{C} - T_{m}) = m_{K} S_{K} (T_{m} - T_{K})$

$0.3 \times 2100 \times (50 - T_{m}) =$ $0.5 \times 2000 \times (T_{m} - 10)$

$\Rightarrow T_{m} = {25.46}^{\circ} C$ .

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