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 3em °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) (a) 15 (b) 10 (c) 6.67 (d) 11.6

Answer: (b). Ice gains: 10 × 2100 × 10 + 10 × 3.36 × 10⁵ + 10 × 4200 T = 3.57 × 10⁶ + 42000 T Water loses: 100 × 4200 (25 - T) = 420000(25 - T) 1.05 × 10⁷ - 420000T = 3.57 × 10⁶ + 42000T ⇒ T = 15°C Decrease = 25 - 15 = 10°C So the answer is (b) 10.

10 kg of ice at -10°C is added to 100 kg of water to lower its… | JEE Main 2026 (28 Jan, Shift 1)

Q9JEE Main 2026 · 28 JAN · SHIFT 1Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat

10 kg of ice at $-10^{\circ}\text{C}$ is added to 100 kg of water to lower its temperature from 25 $^{\circ}\text{C}$ . Consider no heat exchange to surroundings. The decrement to the temperature of water is $\underline{\hspace{3em}}\ {}^{\circ}\text{C}$ . (specific heat of ice $= 2100\text{ J/kg.}{}^{\circ}\text{C}$ , specific heat of water $= 4200\text{ J/kg.}{}^{\circ}\text{C}$ , latent heat of fusion of ice $= 3.36 \times 10^{5}\text{ J/kg}$ )

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Answer: (b)

Ice gains: $10 \times 2100 \times 10 + 10 \times 3.36 \times 10^5 + 10 \times 4200\,T = 3.57 \times 10^6 + 42000\,T$
Water loses: $100 \times 4200 (25 - T) = 420000(25 - T)$
$1.05 \times 10^7 - 420000T = 3.57 \times 10^6 + 42000T \Rightarrow T = 15^{\circ}\text{C}$
Decrease $= 25 - 15 = 10^{\circ}\text{C}$
So the answer is (b) 10.

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