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 =3.36 10⁵ ~J / Kg )

Step-by-step solution

Heat available from water cooling 25°C to 0°C: Q_ available = 100 4200 25 = 10.5 10^6 J. Heat to warm ice from -10°C to 0°C: Q₁ = 10 2100 10 = 0.21 10^6 J. Heat to melt ice at 0°C: Q₂ = 10 3.36 10^5 = 3.36 10^6 J. Total needed: 0.21 + 3.36 = 3.57 10^6 J. Since available heat exceeds required heat, all ice melts. Remaining heat: 10.5 - 3.57 = 6.93 10^6 J warms 110 kg water: T = 6.93 10^6 110 4200 = 15 °C. Final temperature: 15°C. Decrement in water temperature: 25 - 15 = 10 °C.