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10-1.Thermometry, Thermal Expansion and Calorimetry
medium
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
A
$\frac{{50{M_2}}}{{{M_1}}} - 5$
B
$\frac{{5{M_2}}}{{{M_1}}} - 5$
C
$\frac{{50{M_2}}}{{{M_1}}}$
D
$\frac{{5{M_1}}}{{{M_2}}} - 50$
(JEE MAIN-2019)
Solution
$Heat\,lost=Heat\,gain$
$ \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$
Standard 11
Physics
