JEE MAIN - Physics (2017 - 8th April Morning Slot - No. 18)
In an experiment, a sphere of aluminium of mass 0.20 kg is heated upto 150oC. Immediately, it is put into water of volume 150 cc at 27oC kept in a calorimeter of water equivalent to 0.025 kg. Final temperature of the system is 40oC. The specific heat of aluminium is : (take 4.2 Joule = 1 calorie)
378 J/kg $$-$$oC
315 J/kg $$-$$oC
476 J/kg $$-$$oC
434 J/kg $$-$$oC
Explanation
Let specific heat of aluminium = S,
As we know from principle of calorimetry,
Qgiven = Qused
$$\therefore\,\,\,$$ 0.2 $$ \times $$ S $$ \times $$ (150 $$-$$ 40) =
150 $$ \times $$ 1 $$ \times $$ (40 $$-$$ 27) + 25 $$ \times $$ (40$$-$$27)
$$ \Rightarrow $$$$\,\,\,$$ 0.2 $$ \times $$ S $$ \times $$ 110 = 150 $$ \times $$ 13 + 25 $$ \times $$ 13
$$ \Rightarrow $$$$\,\,\,$$ S = 434 J/kg - oC
As we know from principle of calorimetry,
Qgiven = Qused
$$\therefore\,\,\,$$ 0.2 $$ \times $$ S $$ \times $$ (150 $$-$$ 40) =
150 $$ \times $$ 1 $$ \times $$ (40 $$-$$ 27) + 25 $$ \times $$ (40$$-$$27)
$$ \Rightarrow $$$$\,\,\,$$ 0.2 $$ \times $$ S $$ \times $$ 110 = 150 $$ \times $$ 13 + 25 $$ \times $$ 13
$$ \Rightarrow $$$$\,\,\,$$ S = 434 J/kg - oC
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