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Q. 11.14 In an experiment on the specific heat of a metal, a 0.20\; kg  block of the metal at 150 °C is dropped in a copper calorimeter (of water equivalent0.025\; kg) containing 150\; cm^{3} of water at 27^{\circ}C. The final temperature is 40^{\circ}C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal?

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Let the specific heat of the metal be C.

Mass of metal block m = 200 g

Initial Temperature of metal block = 150 oC

Final Temperature of metal block = 40 oC

The heat released by the block is

\\\Delta Q=mc\Delta T\\ \Delta Q=200\times c\times (150-40)\\ \Delta Q=22000c

Initial Temperature of the calorimeter and water = 27 oC

Final Temperature of the calorimeter and water = 40 oC

Amount of water = 150 cm

Mass of water = 150 g

Water equivalent of calorimeter = 25 g

Specific heat of water = 4.186 J g-1 K-1

Heat absorbed by the Calorimeter and water is \\\Delta Q'

\\\Delta Q'=(150+25)\times 4.186\times (40-27)\\ \Delta Q'=9523.15J

The heat absorbed by the Calorimeter and water is equal to the heat released by the block

\\\Delta Q=\Delta Q'\\ 22000c=9523.15\\ c=0.433\ J\ g^{-1}\ K^{-1}

The above value would be lesser than the actual value since some heat must have been lost to the surroundings as well which we haven't accounted for.

 

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Sayak

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