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 ______ ° C .

Question

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 ______ $° C .$

$(specific heat of ice = 2100 J/kg ° C, specific heat of water = 4200 J/kg ° C, latent heat of fusion of ice = 3.36 \times 10^{5} J/kg)$ [2026]

Smart Achievers · Accepted Answer

(2)

$10 \times 3.36 \times 10^{5} + 10 \times 2100 \times 10 + 10 \times 4200 \times (T - 0)$

$= 100 \times 4200 \times (25 - T)$

$\Rightarrow T = 15 ° C$

$Δ T = 25 - 15 = 10 ° C$

Solution

1 11.6

2 10

3 15

4 6.67

Ans.
(2)

$10 \times 3.36 \times 10^{5} + 10 \times 2100 \times 10 + 10 \times 4200 \times (T - 0)$

$= 100 \times 4200 \times (25 - T)$

$\Rightarrow T = 15 ° C$

$Δ T = 25 - 15 = 10 ° C$

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