CalorimetryMedium
Question
10 kg of ice at $- 10^{\circ}C$ 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 $= 2100\text{ }J/Kg.\ ^{\circ}C$, specific heat of water $= 4200\text{ }J/Kg.\ ^{\circ}C$, latent heat of fusion of ice $= 3.36 \times 10^{5}\text{ }J/Kg$ )
Options
A.10
B.15
C.6.67
D.11.6
Solution
$10 \times 3.36 \times 10^{5} + 10 \times 2100 \times 10 + 10 \times 4200 \times (T - 0)$
$${= 100 \times 4200 \times (25 - T) }{\Rightarrow T = 15^{\circ}C }{\Delta T = 25 - 15 = 10^{\circ}C}$$
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