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Calorimeter

10 kg of hot ...

Question

$10 k g$ of hot water in a bucket at $70^{\circ} C$ is cooled for taking a bath adding to it $20 k g$ water at $20^{\circ} C$ . What is the temperature of the mixture?(Neglect the thermal capacity of the bucket)

A

{30.67}^{\circ} C

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B

{36.67}^{\circ} C

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C

{60.67}^{\circ} C

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D

{46.67}^{\circ} C

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Solution

The correct option is B ${36.67}^{\circ} C$
By the energy conservation, thermal energy lost by hot water $=$ thermal energy gained by cold water
If $T_{f}$ be the final temperature of mixture, $m_{h} S (T_{h} - T_{f}) = m_{c} S (T_{f} - T_{c})$ where $S =$ specific of water
or $m_{h} T_{h} - m_{h} T_{f} = m_{c} T_{f} - m_{c} T_{c}$
or $m_{h} T_{h} + m_{c} T_{c} = T_{f} (m_{h} + m_{c})$
or $T_{f} = \frac{m_{h} T_{h} + m_{c} T_{c}}{m_{h} + m_{c}} = \frac{[(10) (70) + (20) (20)]}{10 + 20} = {36.67}^{o} C$

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