20 g of ice at -10^circC is dropped into 80 g…

20 g of ice at -10^circC is dropped into 80 g of water at 40^circC in an insulated vessel of negligible heat capacity. Specific heat of ice =0.5 cal…

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20\ \mathrm{g}

of ice at

-10^\circ\mathrm{C}

is dropped into

80\ \mathrm{g}

of water at

40^\circ\mathrm{C}

in an insulated vessel of negligible heat capacity. Specific heat of ice

=0.5\ \mathrm{cal\,g^{-1}\,^\circ C^{-1}}

, latent heat of fusion

=80\ \mathrm{cal\,g^{-1}}

and specific heat of water

=1\ \mathrm{cal\,g^{-1}\,^\circ C^{-1}}

. The final temperature of the mixture is

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Answer: (C) $15^\circ\mathrm{C}$

Heat the water can give while cooling to

0^\circ\mathrm{C}

80\times1\times40=3200\ \mathrm{cal}

. Heat needed to warm the ice to

0^\circ\mathrm{C}

20\times0.5\times10=100\ \mathrm{cal}

, and to melt it is

20\times80=1600\ \mathrm{cal}

, a total of

1700\ \mathrm{cal}<3200\ \mathrm{cal}

. So all the ice melts and the mixture ends above

0^\circ\mathrm{C}

. The surplus

1500\ \mathrm{cal}

heats the total

100\ \mathrm{g}

of water:

T=\dfrac{1500}{100\times1}=15^\circ\mathrm{C}

. Always compare the heat available with the heat required before assuming a final temperature.

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