JEE Mains · Physics · STD 11 - 10.1, thermonetry,thermal expansion and calorimetry
- JEE Mains 2019
- Medium
- MCQ
When \(M_1\) gram of ice at \(-10\,^oC\) (specific heat \(= 0.5\, cal\, g^{-1}\,^oC^{-1}\)) is added to \(M_2\) gram of water at \(50\,^oC\), finally no ice is left and the water is at \(0\,^oC\). The value of latent heat of ice, in \(cal\, g^{-1}\) is
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Answer & Solution
Correct Answer
(A) \(\frac{{50{M_2}}}{{{M_1}}} - 5\)
Step-by-step Solution
Detailed explanation
\(Heat\,lost=Heat\,gain\)
\( \Rightarrow \,{M_2} \times 1 \times 50 = {M_1} \times 0.5 \times 10 + {M_1}.{L_f}\)
\( \Rightarrow \,\,{L_f} = \frac{{500{M_2} - 5{M_1}}}{{{M_1}}}\)
\( = \frac{{500{M_2}}}{{{M_1}}} - 5\)
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