Thermal Properties of Matter

JEE Main level, single correct, medium

83 of 126

Question 83 of 126

A mass of 50 g50 \mathrm{~g} of water in a closed vessel, with surroundings at a constant temperature takes 2 minutes to cool from 30∘C30^{\circ} \mathrm{C} to 25∘C25^{\circ} \mathrm{C}. A mass of 100 g100 \mathrm{~g} of another liquid in an identical vessel with identical surroundings takes the same time to cool from 30∘C30^{\circ} \mathrm{C} to 25∘C25^{\circ} \mathrm{C}. The specific heat of the liquid is :
(The water equivalent of the vessel is 30 g30 \mathrm{~g}.)

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Answer: (D) 0.5kcal/kg0.5 \mathrm{kcal} / \mathrm{kg}

As the surrounding is identical, vessel is identical time taken to cool both water and liquid (from 30∘C30^{\circ} \mathrm{C} to 25∘C25^{\circ} \mathrm{C} ) is same 2 minutes, therefore
(dQdt)water =(dQdt)liquid \left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\text {water }}=\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\text {liquid }}
or, (mWCW+W)ΔTt=(mℓCℓ+W)ΔTt\frac{\left(\mathrm{m}_{\mathrm{W}} \mathrm{C}_{\mathrm{W}}+\mathrm{W}\right) \Delta \mathrm{T}}{\mathrm{t}}=\frac{\left(\mathrm{m}_{\ell} \mathrm{C}_{\ell}+\mathrm{W}\right) \Delta \mathrm{T}}{\mathrm{t}}
(W=(\mathrm{W}= water equivalent of the vessel ))
or, mwCw=mℓCℓ\mathrm{m}_{\mathrm{w}} \mathrm{C}_{\mathrm{w}}=\mathrm{m}_{\ell} \mathrm{C}_{\ell}
∴\therefore Specific heat of liquid, Cℓ=mWCWmℓ\mathrm{C}_{\ell}=\frac{\mathrm{m}_{\mathrm{W}} \mathrm{C}_{\mathrm{W}}}{\mathrm{m}_{\ell}}
=50×1100=0.5kcal/kg=\frac{50 \times 1}{100}=0.5 \mathrm{kcal} / \mathrm{kg}

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