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A sphere of 0.047 kg aluminium 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 kgcopper calorimeter containing 0.25 kg water at 20^(@) C . The temperature of water rises and attains a steady state at 23 ^(@) C. Calculate the specific heat capacity of aluminium. |
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Answer» Solution :In solving this example, we shall use the fact that at a steady STATE, heat given by an aluminium sphere will be equal to the heat absorbed by the water and calorimeter. Mass of aluminium sphere`(m_(1)) = 0.047 ` kgInitial temperature of aluminium sphere`= 100 ^(@) C` Final temperature`= 23^(@) C` Change in temperature `(DELTA T) = (100 ^(@) C - 23 ^(@) C) = 77 ^(@) C` LET specific heat capacity of aluminium be `s_(Al)`. Theamount of heat lost by the aluminium sphere = `m_(1) s_(Al) Delta T = 0.047 kg xx s_(Al) xx 77 ^(@) C` Massof water `(m_(2)) = 0.25 kg` Mass of calorimeter `(m_(3)) = 0.14 kg` Initialtemperature of water and calorimeter = `20^(@) C` Final temperature of the mixture ` = 23 ^(@) C` Change in temperature `(Delta T_(2)) = 23 ^(@) C - 20 ^(@) C = 3 ^(@) C` Specific heat capacity of copper calorimeter ` = 0.386 xx 10^(3) J kg^(-1) K^(-1)` Specific heat capacity of copper calorimeter ` = 0.386 xx 10^(3) J kg^(-1) K^(-1)` The amount of heat GAINED by water and calorimeter ` = m_(2) s_(w) Delta T_(2) + m_(3) s_(cu) Delta T_(2)` `= (m_(2) s_(w) + m_(3) s_(su)) (Delta T_(2))` ` = (0.25 kg xx 4.18 xx 10^(3) J kg ^(-1) K^(-1) + 0.14 kg xx 0.386 xx 10^(3) J kg^(-1) K^(-1) (23 ^(@) C - 20 ^(@) C)` In the steady state heat lost by the aluminium sphere = heat gained by water `+ ` heat gained by calorimeter. So, `0.047 kg xx s_(Al) xx 77^(@) C` ` = (0.25 kg xx 4.18 xx 10^(3) J kg^(-1) K^(-1) + 0.14 kg xx 0.386 xx 10^(3) J kg^(-1) K^(-1)) ( 3^(@) C)` ` s_(Al) = 0.911 kJ kg^(-1) K^(-1)` |
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