Physics · 2020
JEE Main · 4 September 2020, Shift 1 · Q7
The specific heat of water = 4200 J kg^-1 K^-1 and the latent heat of ice =3.4 × 10^5 J kg^-1. 100 grams of ice at 0^° C is placed in 200 g of water…
The specific heat of water = $\displaystyle 4200 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$ and the latent heat of ice $\displaystyle =3.4 \times 10^5 \mathrm{~J} \mathrm{~kg}^{-1}$. $\displaystyle 100$ grams of ice at $\displaystyle 0^{\circ} \mathrm{C}$ is placed in $\displaystyle 200$ g of water at $\displaystyle 25^{\circ} \mathrm{C}$. The amount of ice that will melt as the temperature of water reaches $\displaystyle 0$°C is close to ( in grams) :
Official answer
(2)
$\displaystyle 61.7$
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JEE Main 2020 Physics question, with the answer from NTA’s final answer key. Where our answers come from.