Q61 / 75 · JEE Main 2026 (28 January Shift 1) · Physics
10 kg of ice at $-10^{\circ} \mathrm{C}$ is added to 100 kg of water to lower its temperature from 25 ${ }^{\circ} \mathrm{C}$. Consider no heat exchange to surroundings. The decrement to the temperature of water is $\_\_\_\_$ ${ }^{\circ} \mathrm{C}$.
(specific heat of ice $=2100 \mathrm{~J} / \mathrm{Kg}.{ }^{\circ} \mathrm{C}$, specific heat of water $=4200 \mathrm{~J} / \mathrm{Kg}.{ }^{\circ} \mathrm{C}$, latent heat of fusion of ice $=3.36 \times 10^{5} \mathrm{~J} / \mathrm{Kg}$)
(specific heat of ice $=2100 \mathrm{~J} / \mathrm{Kg}.{ }^{\circ} \mathrm{C}$, specific heat of water $=4200 \mathrm{~J} / \mathrm{Kg}.{ }^{\circ} \mathrm{C}$, latent heat of fusion of ice $=3.36 \times 10^{5} \mathrm{~J} / \mathrm{Kg}$)
- A6.67
- B11.6
- C15
- Correct answerD10
Solution
Heat available from water cooling 25°C to 0°C: $Q_{available} = 100 \times 4200 \times 25 = 10.5 \times 10^6$ J.
Heat to warm ice from -10°C to 0°C: $Q_1 = 10 \times 2100 \times 10 = 0.21 \times 10^6$ J.
Heat to melt ice at 0°C: $Q_2 = 10 \times 3.36 \times 10^5 = 3.36 \times 10^6$ J.
Total needed: $0.21 + 3.36 = 3.57 \times 10^6$ J.
Since available heat exceeds required heat, all ice melts. Remaining heat: $10.5 - 3.57 = 6.93 \times 10^6$ J
warms 110 kg water: $\Delta T = \frac{6.93 \times 10^6}{110 \times 4200} = 15$ °C.
Final temperature: 15°C.
Decrement in water temperature: $25 - 15 = 10$ °C.
Heat to warm ice from -10°C to 0°C: $Q_1 = 10 \times 2100 \times 10 = 0.21 \times 10^6$ J.
Heat to melt ice at 0°C: $Q_2 = 10 \times 3.36 \times 10^5 = 3.36 \times 10^6$ J.
Total needed: $0.21 + 3.36 = 3.57 \times 10^6$ J.
Since available heat exceeds required heat, all ice melts. Remaining heat: $10.5 - 3.57 = 6.93 \times 10^6$ J
warms 110 kg water: $\Delta T = \frac{6.93 \times 10^6}{110 \times 4200} = 15$ °C.
Final temperature: 15°C.
Decrement in water temperature: $25 - 15 = 10$ °C.
Hence, the correct answer is option: 'D'.