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Physics · 2019

JEE Main · 11 January 2019, Shift 1 · Q9

Ice at -20^° C is added to 50 g of water at 40^° C. When the temperature of the mixture reaches 0°C, it is found that 20 g of ice is still unmelted.…

Ice at $\displaystyle -20^{\circ} \mathrm{C}$ is added to $\displaystyle 50$ g of water at $\displaystyle 40^{\circ} \mathrm{C}$. When the temperature of the mixture reaches $\displaystyle 0$°C, it is found that $\displaystyle 20$ g of ice is still unmelted. The amount of ice added to the water was close to ( Specific heat of water $\displaystyle =4.2 \mathrm{~J} / \mathrm{g} /{ }^{\circ} \mathrm{C}$ Specific heat of Ice = $\displaystyle 2.1$ J/g/°C Heat of fusion of water at $\displaystyle 0^{\circ} \mathrm{C}=334 \mathrm{~J} / \mathrm{g}$ )
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JEE Main 2019 Physics question, with the answer from NTA’s final answer key. Where our answers come from.