A 16 gram sample of water at 273 K is cooled so that it becomes a completely solid ice cube at 273 K. How much heat was released by the sample of water to form this ice cube?

16 J

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4368 J

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18258 J

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350 J

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5334 J

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Solution

The correct option is D 5334 J
The enthalpy of fusion of ice $Δ H_{f u s} = 6000 J / m o l = \frac{6000}{18} J / g = 333.33 J / g$
When 16 gram water sample is cooled at 273 K so that it becomes a completely solid ice cube at 273 K, heat released $q = 333.33 J / g \times 16 g \approx 5334 J$

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100 grams

0^{\circ} C

27^{\circ} C

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