In an experiment to determine 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 equivalent `0.025 kg`) containing `150 cm^3` of water at `27 .^(@) C`. The final temperature is `40.^(@) C`. The specific heat of the metal is.

Options

A. `434.3 J//kg- .^(@)C`
B. `222.6 J//kg-.^(@)C`
C. `167.3 J//kg- .^(@)C`
D. `712.5 J//kg-.^(@)C`

Answer

A

Explanation

Let specific heat of the metal is s, the heat lost by metal block `= ms Delta theta`
`= 0.02 s (150 - 40) = 22 s`
Mass of watr in the calorimeter `= rho V`
`= 1000 xx (150 xx 10^(-6)) = 0.15 kg`
Heat gained by water and calorimeter
`m_(w) s Deltha theta + W Delta theta`
`= (0.15 xx 4200 + 0.025 xx 4200) xx (40 - 27)`
`735 xx 13 = 9555 J`
By principle of calorimetry, we have
`22s = 9555`
`:. s = (9555)/(22) = 434.3 J//kg - .^(@)C`