1. Tardigrade
  2. Question
  3. Physics
  4. When M1 gram of ice at -10°C (specific heat = 0.5 cal g-1 ° C-1) is added to M2 gram of water at 50°C, finally no ice is left and the water is at 0°C. The value of latent heat of ice, in cal g-1 is:

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Q. When $M_1$ gram of ice at $-10^{\circ}C$ (specific heat = $0.5 \; cal \; g^{-1 \circ } C^{-1}$) is added to $M_2$ gram of water at $50^{\circ}C$, finally no ice is left and the water is at $0^{\circ}C$.
The value of latent heat of ice, in cal $g^{-1}$ is:

JEE MainJEE Main 2019Thermal Properties of Matter

Solution:

$Heat\, lost = Heat \,gain$
$\Rightarrow \; M_2 \times 1 \times 50 = M_1 \times 0.5 \times 10 + M_1.L_f$
$\Rightarrow \; L_f = \frac{50 M_2 - 5 M_1}{M_1}$
$ = \frac{50 M_2 }{M_1} - 5 $