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InterviewSolution
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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)` |
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Answer» 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)`. |
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