200g of a solid ball at `20^(@)C` is dropped in an equal amount of water at `80^(@)C`. The resulting temperature is `60^(@)C`. This means that specific heat of solid is

Options

A. One fourth of water
B. one half of water
C. twice of water
D. four times of water

Answer

B

Explanation

Heat lost by the water, `Q_(1) = 200 xx s_(w) xx (80-60)`
Heat gained by the solid, `Q_(2) = 200 xx s_(s) xx (60-20)`
According to the principle of calorimetry `Q_(1) = Q_(2)`
`:. 200 xx s_(w) xx (80 -60) = 200 xx s_(w) xx (60-20)`
or `s_(s) = (1)/(2) s_(w)`