10 kg of ice at -10°C is added to 100 kg of water to lower its temperature from 25°C. Consider no heat exchange to surroundings. The decrement to the temperature of water is _____ °C.
(specific heat of ice= 2100 J/Kg.°C, specific heat of water= 4200 J/Kg.°C, latent heat of fusion of ice $$=3.36\times\ 10^5J/Kg$$)

We have 10 kg of ice at -10 °C and 100 kg of water at 25 °C.

The specific heats are $$c_{ice} = 2100$$ J/kg°C and $$c_{water} = 4200$$ J/kg°C, with latent heat of fusion $$L_f = 3.36 \times 10^5$$ J/kg.

First, warming the ice to 0 °C requires $$Q_1 = 10 \times 2100 \times 10 = 210000$$ J, and melting it requires $$Q_2 = 10 \times 336000 = 3360000$$ J, giving a total heat requirement of $$3570000$$ J.

The heat available from cooling 100 kg of water from 25 °C to 0 °C is $$10500000$$ J, which exceeds $$Q_1 + Q_2$$, so all the ice melts.

Let the final temperature of the mixture be $$T$$. Then heat lost by water cooling from 25 °C to $$T$$ equals heat gained by the ice warming and melting, giving

$$100 \times 4200 \times (25-T) = 210000 + 3360000 + 10 \times 4200 \times T.$$

Rearranging gives

$$10500000 - 420000T = 3570000 + 42000T,$$

so

$$6930000 = 462000T$$

and hence

$$T = 15°C.$$

Thus the temperature decrease is $$25 - 15 = 10$$°C. The correct answer is Option 2: 10°C.