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 $=3.36 \times 10^{5} \mathrm{~J} / \mathrm{Kg}$)

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Heat & Thermodynamics

10 kg of ice at

- 1 0^{\circ} C

is added to 100 kg of water to lower its temperature from

2 5^{\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

)

Answer: (A)

The ice at

- 1 0^{\circ} C

must first warm to

0^{\circ} C

, then melt, then warm to final temperature

T

. The water cools from

2 5^{\circ} C

T

. By conservation of energy:

Heat gained by ice = Heat lost by water

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

210000 + 3360000 + 42000 T = 10500000 - 420000 T

462000 T = 6930000

T = 1 5^{\circ} C

Decrement in temperature:

Δ T = 25 - 15 = 1 0^{\circ} C

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Single correct option