JEE Main 2026 — Jan 28, Shift 1 — Physics solution

JEE Main2026Jan 28, Shift 1Physics

10 kg of ice at $- 1 0^{\circ}$ C is added to 100 kg of water to lower its temperature from 25 $^{\circ}$ C. Consider no heat exchange to surroundings. The decrement to the temperature of water is ______ $^{\circ}$ C.

(specific heat of ice = 2100 J/Kg. $^{\circ}$ C, specific heat of water = 4200 J/Kg. $^{\circ}$ C, latent heat of fusion of ice = $3.36 \times 1 0^{5}$ J/Kg)

A
10
B
15
C
6.67
D
11.6

Solution

Heat absorbed by ice (warm $- 10 \to 0$ , melt, then warm $0 \to T$ ) equals heat released by water (cool $25 \to T$ ):

$m_{ice} c_{ice} (10) + m_{ice} L + m_{ice} c_{w} (T) = m_{w} c_{w} (25 - T)$

$10 \cdot 2100 \cdot 10 + 10 \cdot 3.36 \times 1 0^{5} + 10 \cdot 4200 \cdot T = 100 \cdot 4200 \cdot (25 - T)$

$2.1 \times 1 0^{5} + 33.6 \times 1 0^{5} + 42000 T = 4.2 \times 1 0^{5} (25 - T)$

$35.7 \times 1 0^{5} + 4.2 \times 1 0^{4} T = 105 \times 1 0^{5} - 4.2 \times 1 0^{5} T$

$4.62 \times 1 0^{5} T = 69.3 \times 1 0^{5} \Rightarrow T = 1 5^{\circ}$ C.

Decrement to water's temperature $= 25 - 15 = 1 0^{\circ}$ C. Option (1).

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