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JEE Mains · Physics · STD 11 - 10.1, thermonetry,thermal expansion and calorimetry
- JEE Mains 2013
- Hard
- MCQ
A mass of \(50\,g\) of water in a closed vessel, with surroundings at a constant temperature takes \(2\, minutes\) to cool from \(30\,^oC\) to \(25\,^oC\). A mass of \(100\,g\) of another liquid in an identical vessel with identical surroundings takes the same time to cool from \(30\,^oC\) to \(25\,^o C\). The specific heat of the liquid is .......... \(kcal/kg\) (The water equivalent of the vessel is \(30\,g\).)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(0.5\)
Step-by-step Solution
Detailed explanation
As the surrounding is identical, vessel is identical time taken to cool both water and liquid\(\left( {from\,{{30}^ \circ }C\,to\,{{25}^ \circ }C} \right)\) is same \(2\,minutes,\) therefore
\({\left( {\frac{{dQ}}{{dt}}} \right)_{water}} = {\left( {\frac{{dQ}}{{dt}}} \right)_{liquid}}\)
\(or,\frac{{\left( {{m_w}{C_w} + W} \right)\Delta T}}{t} = \frac{{\left( {{m_\ell }{C_\ell } + W} \right)\Delta T}}{t}\)
\(\left( {W = water\,equivalent\,of\,the\,vessel} \right)\)
\(or,\,\,{m_w}{C_w} = {m_\ell }{C_\ell }\)
\(\therefore Specific\,heat\,of\,liquid,\,{C_\ell } = \frac{{{m_W}{C_W}}}{{{m_\ell }}}\)
\( = \frac{{50 \times 1}}{{100}} = 0.5\,kcal/kg\)
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