Question

A sphere of alumininum of mass 0.047 kg placed for sufficient time in a vessel containing boling 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^(@)C` . calculate the specific heat capacity of aluminum. Specific heat capacity of copper = `0.386 xx 10^(3) J kg^(-1) K^(-1)`.
Specific heat capacity of water = `4.18 xx 10^(-3) J kg^(-1) K^(-1)`

Answer 1

Here, `m_(Al) = 0.047 kg` ,
`T_(1) = 100^(@)C , m_(cu) = 0.14 kg` ,
`m_(w) = 0.25 kg , T_(0) = 20^(@)C`
`T_(2) = 23^(@) C , s_(Cu) = 0.386 xx 10^(3) J kg^(-1) K^(-1)`
Heat lost by aluminium,
`Q_(1) = m_(Al)s_(Cu) (T_(1)-T_(2))`
`=0.047 xx s_(Al) xx (100 -23)`
`=0.047 xx s_(Al) xx 77 J`
Heat taken by copper calorimeter and water is
`Q_(2) = m_(Cu) s_(Cu) (T_(2)-T_(0))+m_(w)s_(w)(T_(2)-T_(0))`
`= 0.14 xx (0.386 xx 10^(3)) xx (23 -20)`
`+0.25 xx (4.18 xx 10^(3)) (23-20)`
`= 162.12 + 3135 = 3197.12 J`
In the steady state,haet lost= heat gained
`:. 0.04 xx s_(Al) xx 77 = 3297.12`
or, `s_(Al) = (3297.12)/(0.047 xx 77) = 911 J kg^(-1)K^(-1)`.