A mass of 50 ~g of water in a closed vessel, with surroundings at a constant temperature takes 2 minutes to cool from 30^ C to 25^ C . A mass of 100 ~g of another liquid in an identical vessel with identical

Correct answer: D. 0.5 kcal / kg As the surrounding is identical, vessel is identical time taken to cool both water and liquid (from 30^ ^ C to 25^ ^ C ) is same 2 minutes, therefore ( dQ dt )_ water =( dQ dt )_ liquid or, (m_ W C_ W +W) Δ T t = (m_ C_ +W) Δ T t (W= water equivalent of the vessel ) or, m_ w C_ w =m_ C_ Specific heat of liquid, C_ = m_ W C_ W m_ =(50 × 1)/(100)= 0.5 kcal / kg

JEE Main 201325 AprilOnline

Exam: JEE Main
Year: 2013
Date: 25 April
Mode: Online
Subject: Physics
Chapter: Thermal Properties of Matter

JEE Main 2013 · 25 April · Online · Physics · Thermal Properties of Matter

A mass of 50 ~g of water in a closed vessel, with surroundings at a...

A mass of

50 \mathrm{~g}

of water in a closed vessel, with surroundings at a constant temperature takes 2 minutes to cool from

30^{\circ} \mathrm{C}

25^{\circ} \mathrm{C}

. A mass of

100 \mathrm{~g}

of another liquid in an identical vessel with identical surroundings takes the same time to cool from

30^{\circ} \mathrm{C}

25^{\circ} \mathrm{C}

. The specific heat of the liquid is : (The water equivalent of the vessel is

30 \mathrm{~g}

Options

A $2.0 \mathrm{kcal} / \mathrm{kg}$
B $7 \mathrm{kcal} / \mathrm{kg}$
C $3 \mathrm{kcal} / \mathrm{kg}$
D $0.5 \mathrm{kcal} / \mathrm{kg}$

Correct answer

D. $0.5 \mathrm{kcal} / \mathrm{kg}$

Step-by-step solution

As the surrounding is identical, vessel is identical time taken to cool both water and liquid (from

30^{{}^{\circ}} \mathrm{C}

25^{{}^{\circ}} \mathrm{C}

) is same 2 minutes, therefore

\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\text{water }}=\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\text{liquid }}

or,

\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}}(\mathrm{W}=

water equivalent of the vessel

)

or,

\mathrm{m}_{\mathrm{w}} \mathrm{C}_{\mathrm{w}}=\mathrm{m}_{\ell} \mathrm{C}_{\ell}\therefore

Specific heat of liquid,

\mathrm{C}_{\ell}=\frac{\mathrm{m}_{\mathrm{W}} \mathrm{C}_{\mathrm{W}}}{\mathrm{m}_{\ell}}=\frac{50 \times 1}{100}=

0.5

\mathrm{kcal}

\mathrm{kg}

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