200, g of a solid ball at 20,^oC is dropped in an equal amount of water at 80,^oC . resulting temper

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10-1.Thermometry, Thermal Expansion and Calorimetry

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$200\, g$ of a solid ball at $20\,^oC$ is dropped in an equal amount of water at $80\,^oC$ . The resulting temperature is $60\,^oC$ . This means that specific heat of solid is

A

One fourth of water

B

One half of water

C

Twice of water

D

Four times of water

Solution

$200 \times S_{solid} \times (60 -20) = 200 \times S_w (80 -60)$
Here $S_w = 1 \,cal/gm-^oC$

Standard 11

Physics

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