Search any question & find its solution
Question:
Answered & Verified by Expert
A mass of $50 \mathrm{~g}$ of water in a closed vessel, with surroundings at a constant temperature takes 2 minutes to cool from $30^{\circ} \mathrm{C}$ to $25^{\circ} \mathrm{C}$. A mass of $100 \mathrm{~g}$ of another liquid in an identical vessel with identical surroundings takes the same time to cool from $30^{\circ} \mathrm{C}$ to $25^{\circ} \mathrm{C}$. The specific heat of the liquid is :
(The water equivalent of the vessel is $30 \mathrm{~g}$.)
Options:
(The water equivalent of the vessel is $30 \mathrm{~g}$.)
Solution:
2025 Upvotes
Verified Answer
The correct answer is:
$0.5 \mathrm{kcal} / \mathrm{kg}$
$0.5 \mathrm{kcal} / \mathrm{kg}$
As the surrounding is identical, vessel is identical time taken to cool both water and liquid (from $30^{\circ} \mathrm{C}$ to $25^{\circ} \mathrm{C}$ ) 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, $\mathrm{C}_{\ell}=\frac{\mathrm{m}_{\mathrm{W}} \mathrm{C}_{\mathrm{W}}}{\mathrm{m}_{\ell}}$
$$
=\frac{50 \times 1}{100}=0.5 \mathrm{kcal} / \mathrm{kg}
$$
$$
\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\text {water }}=\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\text {liquid }}
$$
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, $\mathrm{C}_{\ell}=\frac{\mathrm{m}_{\mathrm{W}} \mathrm{C}_{\mathrm{W}}}{\mathrm{m}_{\ell}}$
$$
=\frac{50 \times 1}{100}=0.5 \mathrm{kcal} / \mathrm{kg}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.