A sphere of aluminium of 0.047 kg placed for sufficient time in a vessel containing boiling water, so that 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" "^(@)C. Calculate the specific heat capacity of aluminium.

A sphere of aluminium of 0.047 kg placed for sufficient time in a vess

1.	A sphere of aluminium of 0.047 kg placed for sufficient time in a vessel containing boiling water, so that 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" "^(@)C. Calculate the specific heat capacity of aluminium.
Answer» Solution :Mass of sphere `m_(1)=0.047` KG Initial temperature of sphere `T_(1)=100^(@)C` Final temperature of sphere `T_(2)=23^(@)C` Change in temperature `DeltaT_(1)=T_(1)-T_(2)` `=(100-23)" "^(@)C` `=77^(@)C` Sphere specific heat of sphere `=S_(1)` Heat lost by aluminum sphere, `Q_(1)=m_(1)S_(1)DeltaT_(1)` `=0.047xxS_(1)xx77` `:.Q_(1)=3.619S_(1)`. . .(1) Now, mass of water `m_(2)=0.25` kg Mass of calorimeter `m_(3)=0.14` kg Specific heat of water, `S_(2)=4.18xx10^(3)" Jkg"^(-1)K^(-1)` Specific heat of copper calorimeter, `S_(3)=0.386xx10^(3)" Jkg"^(-1)K^(-1)` Heat gained by water and calorimeter, `Q_(2)=m_(2)S_(2)DeltaT_(2)+m_(3)S_(3)DeltaT_(2)` `=0.25xx4.18xx10^(3)xx3+0.14xx0.386xx10^(3)xx3` `=3.135xx10^(3)+0.16212xx10^(3)` `:.Q_(2)=3.29712xx10^(3)`. . .(2) For STEADY state, {Heat lost by sphere} = {Heat gained by water} + {Heat gained by calorimeter} `Q_(1)=Q_(2)` `3.619S_(1)=3.29712xx10^(3)` [from equation (1) and (2)] `:.S_(1)=(3.29712xx10^(3))/(3.619)=0.911xx10^(3)` `:.S_(1)=0.911" kJ kg"^(-1)=911" Jkg"^(-1)K^(-1)` (specific heat of aluminium)