A sphere of aluminium of 0.047 kg is placed for sufficient time in a vessel containing boiling water, so that the sphere is at `100^(@)` C. It is then immediately transferred to 0.14 kg copper calorimeter containing 0.25 kg of water at `20^(@)` C. The temperature of water rises and attains a steady state at `23^(@)`Calculate the specific heat capacity of aluminium. (Give specific heat of copper `=0.386xx10^(3)Jkg^(-1)K^(-1))`.

Explanation

Given, mass of aluminium sphere,
m = 0.047 kg
Initial temperature of sphere = `100^@C`
Final temperature of sphere = `23^@C `
Fall in temperature of sphere, `Delta T = 100 - 23 = 77^@C`
If `C_(Al)` is the specific heat of aluminium, then heat lost by the aluminium sphere=`mC_(Al) Delta T = 0.047 xx C_(Al) xx 77`
Mass of water, `m_w = 0.25 kg`
Mass of calorimeter, `m_c=0.14 kg`
Initial temperature of water and calorimeter, `T_1 = 20^@C`
Final temperature of water and calorimeter, `T_2 = 23^@C `
Rise in temperature of water and calorimeter,
`Delta T. = T_2-T_1 = 23- 20 = 3^@C`
Specific heat of water, `C_w = 4.18 10^3J kg^(-1) K^(-1) `
Specific heat of copper, `C_(cu) = 0.386 xx 10^3 J kg^(-1) K^(-1)`
Heat gained by (water + calorimeter) is
`H=m_w C_w Delta T. + m_c C_(cu) Delta T.`
`= (0.25 xx 4.18 xx 10+ 0.14 xx 0.386 xx 10^2) xx 3`
In the steady state
Heat lost by sphere = Heat gained by water + calorimeter
`implies 0.047 xx C_(Al) xx 77 = (0.25 xx 4.18 xx 10^3 + 0.14 xx 0.386 xx 1063) xx 3`
` C_(Al) = 0.911 xx 10^8 kJ kg^(-1) K^(-1)`