A calorimeter of mass m contains an equal mass of water in it- The temperature of the water and calorimeter is t2-0- C- A block of ice of mass m and temperature t30B- C 1- C 2 t 2- C 3- t 3- L -0C- C 1- C 2 t 2- C 3- t 3- L -0D- C 1- C 2 t 2- C 3- t 3- L -0

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Calorimetry

A calorimeter...

Question

A calorimeter of mass m contains an equal mass of water in it. The temperature of the water and calorimeter is $t_{2} > 0^{\circ} C$ . A block of ice of mass m and temperature $t_{3} < 0^{\circ} C$ is gently dropped into the calorimeter. Let $C_{1}, C_{2}$ and $C_{3}$ be the specific heats of material of calorimeter, water and ice respectively and L be the Latent heat of ice. The whole mixture in the calorimeter becomes water, if

(C_{1} + C_{2}) t_{2} - C_{3} | t_{3} | + L > 0

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(C_{1} + C_{2}) t_{2} + C_{3} | t_{3} | + L > 0

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(C_{1} + C_{2}) t_{2} - C_{3} | t_{3} | - L > 0

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(C_{1} + C_{2}) t_{2} + C_{3} | t_{3} | - L > 0

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Solution

The correct option is C $(C_{1} + C_{2}) t_{2} - C_{3} | t_{3} | - L > 0$
For whole water, heat supplied by ice = $m C_{3} t_{3} + m L$ .
That will be gained by calorimeter and water.
The heat absorbed by the calorimeter and water can be given as $m C_{1} t_{2} + m C_{2} t_{2}$ .
For converting whole mixture to water,
$m C_{1} t_{2} + m C_{2} t_{2} > m C_{3} t_{3} + m L$
Hence, $(C_{1} + C_{2}) t_{2} - C_{3} | t_{3} | - L > 0$

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