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.

Explanation

Here mass of metal, m = 0.20 kg =200 g
Fall in temperature fo metal ` Delta T =150 - 40 =110^@C`
If s is specific heat of the metal, then heat lost by the metal, `Delta Q = m s Delta T = 200 xx s xx 100 …. (i)`
Volume of water ` =150^@c.c :.` Mass of water, `m' =150 g`
Water equivalent of calorimeter, `w = 0.225kg =25 g`
Rise in temp. of water and calorimeter, `Delta T' = 40 - 27 =13^@C`
Heat gained by water and calorimeter, `Delta Q' = (m' +w) Delta T' = (150 + 25) xx 13 = 175 xx13` ...... (ii)
As, `Delta Q = DeltaQ'`
` :.` From (i) and (ii)
`200 xx sxx 110 = 175 xx 13 or s = (17xx13)/(200 xx110) = 0.10341 cal g^(-1) .^@C^(-1) = 0.43 J g^(-1) K^(-1)`
It some heat is lost to the surrounding, value of s so Obtained will be less the actual value of s.