In an experiment a sphere of aluminium of mass 0.20 kg is heated upto `150^(@)`C. Immediately, it is put into water of volume 150 cc at `27^(@)C` kept in a calorimeter of water equivalent to 0.025 kg. Final temperature of the system is `40^(@)C`. The specific heat of aluminium is (take 4.2 Joule = 1 calorie)

Options

A. `378 J// kg ""^@C `
B. `315 J//kg ""^@C`
C. `476 J//kg// ""^@C`
D. `434 J//kg// ""^@C`

Answer

D

Explanation

`T_1 = 150^@C , T_2 = 27^@C , T = 40^@C, M_(Al) = 0.20 kg , M_w = 0.25 kg `
Vol. of water in calorimeter = 150 cc =`150 xx 10^(-6) m^3`
Mass of water in calorimeter,
`M_(wc) = (150 xx 10^(-6) ) xx 10^3 kg = 150 xx 10^(-3) kg `
Heat lost by sphere = Heat gained by water in calorimeter
`M_(Al) C_(Al) (150 - 40) = MC_w (40 - 27)`
` 0.2 C_(Al) (110) =(M_(wc) + M_(w) ) C_w (13)`
` 22C_(Al) =(0.15 + 0.025)4.2 xx 10^(-3) (13)`
`C_(Al) = 0.4343 xx 10^(-3) = 434.3 J kg^(-1) C^(-1)`