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 wate

D

Four times of water

Solution

$200 \times S_{\text {solid }} \times(60-20)=200 \times S_{w}(80-60)$

Here $S_{w}=1$ cal/gm $-^{\circ} \mathrm{C}$

Standard 11

Physics

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