Physics · 2019
JEE Main · 12 April 2019, Shift 1 · Q9
When M_1 gram of ice at -10^° C (specific heat =0.5 cal g^-1^° C^-1 ) is added to M_2 gram of water at 50^° C, finally no ice is left and the water…
When $\displaystyle \mathrm{M}_1$ gram of ice at $\displaystyle -10^{\circ} \mathrm{C}$ (specific heat $\displaystyle =0.5 \mathrm{cal} \mathrm{g}^{-1}{ }^{\circ} \mathrm{C}^{-1}$ ) is added to $\displaystyle \mathrm{M}_2$ gram of water at $\displaystyle 50^{\circ} \mathrm{C}$, finally no ice is left and the water is at $\displaystyle 0^{\circ} \mathrm{C}$. The value of latent heat of ice, in cal $\displaystyle \mathrm{g}^{-1}$ is :
Official answer
(4)
$\displaystyle \frac{50 \mathrm{M}_2}{\mathrm{M}_1}-5$
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JEE Main 2019 Physics question, with the answer from NTA’s final answer key. Where our answers come from.