Thermal Properties of Matter

JEE Main level, single correct, medium

49 of 126

Question 49 of 126

10 kg of ice at −10∘C-10^{\circ} \mathrm{C} is added to 100 kg of water to lower its temperature from 25 ∘C{ }^{\circ} \mathrm{C}. Consider no heat exchange to surroundings. The decrement to the temperature of water is ____\_\_\_\_ ∘C{ }^{\circ} \mathrm{C}.
(specific heat of ice
=2100 J/Kg.∘C=2100 \mathrm{~J} / \mathrm{Kg}.{ }^{\circ} \mathrm{C}, specific heat of water =4200 J/Kg.∘C=4200 \mathrm{~J} / \mathrm{Kg}.{ }^{\circ} \mathrm{C}, latent heat of fusion of ice =3.36×105 J/Kg=3.36 \times 10^{5} \mathrm{~J} / \mathrm{Kg})

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Answer: (D) 10

Heat available from water cooling 25°C to 0°C: Qavailable=100×4200×25=10.5×106Q_{available} = 100 \times 4200 \times 25 = 10.5 \times 10^6 J.
Heat to warm ice from -10°C to 0°C:
Q1=10×2100×10=0.21×106Q_1 = 10 \times 2100 \times 10 = 0.21 \times 10^6 J.
Heat to melt ice at 0°C:
Q2=10×3.36×105=3.36×106Q_2 = 10 \times 3.36 \times 10^5 = 3.36 \times 10^6 J.
Total needed:
0.21+3.36=3.57×1060.21 + 3.36 = 3.57 \times 10^6 J.
Since available heat exceeds required heat, all ice melts. Remaining heat:
10.5−3.57=6.93×10610.5 - 3.57 = 6.93 \times 10^6 J
warms 110 kg water:
ΔT=6.93×106110×4200=15\Delta T = \frac{6.93 \times 10^6}{110 \times 4200} = 15 °C.
Final temperature: 15°C.

Decrement in water temperature:
25−15=1025 - 15 = 10 °C.

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