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  4. Ice at -20° C os added tp 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. The amount of ice added to the water was close to (Specific heat of water = 4.2 J/g/°C) Specific heat of Ice = 2.1 J/g/°C Heat of fusion of water at 0°C = 334 J/g)

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Q. Ice at $-20^{\circ} C$ os added tp $50\, g$ of water at $40^{\circ}C$. When the temperature of the mixture reaches $0^{\circ}C$, it is found that $20\, g$ of ice is still unmelted. The amount of ice added to the water was close to
(Specific heat of water = $4.2 \; J/g/^{\circ}C$)
Specific heat of Ice = $2.1 \; J/g/^{\circ}C$
Heat of fusion of water at $0^{\circ}C = 334 \; J/g$)

JEE MainJEE Main 2019Thermal Properties of Matter

Solution:

Let amount of ice is m gm.
According to principal of calorimeter heat taken by ice = heat given by water
$\therefore \; 20 \times 2.1 \times m + (m -20 ) \times 334$
$ = 50 \times 4.2 \times 40$
376 m = 8400 + 6680
$m = 40.1$