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. The amount of ice added to the water was close to (Specific heat of

JEE Main 201911 JanuaryMorning Shift

JEE Main 2019 · 11 January · Morning Shift · Physics · Thermal Properties of Matter

Ice at -20^ C is added to 50 ~g of water at 40^ C , When the...

Ice at

-20^{\circ} \mathrm{C}

is added to

50 \mathrm{~g}

of water at

40^{\circ} \mathrm{C}

, When the temperature of the mixture reaches

0^{\circ} \mathrm{C},

it is found that 20

\mathrm{g}

of ice is still unmelted. The amount of ice added to the water was close to (Specific heat of water

=4.2 \mathrm{~J} / \mathrm{g} /{ }^{\circ} \mathrm{C}

Specific heat of Ice

=2.1 \mathrm{~J} / \mathrm{g} /{ }^{\circ} \mathrm{C}

Heat of fusion of water at

\left.0^{\circ} \mathrm{C}=334 \mathrm{~J} / \mathrm{g}\right)

D. $40 \mathrm{~g}$

Let m gram of ice is added. From principal of calorimeter heat gained (by ice) = heat lost (by water)

\begin{array}{l} \therefore 20 \times 2.1 \times \mathrm{m}+(\mathrm{m}-20) \times 334 \\ =50 \times 4.2 \times 40 \\ 376 \mathrm{~m}=8400+6680 \\ \mathrm{~m}=40.1 \end{array}