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How many kilocalories must be removed from 920 grams of water at a temperature of zero degrees Celsius to transform it into ice at zero degrees Celsius? A)2760 B)497000 C)73.3

Sagot :

Given:

The mass of the water, m=920 g

The temperature of the water, T=0 °C

To find:

The heat required to be removed to transform water to ice.

Explanation:

The latent heat of the fusion of ice is L=79.7 Cal/g.

The latent heat of fusion is the heat that is needed to be removed from the water to convert it into ice.

The amount of the heat that is needed to be removed is given by,

[tex]Q=mL[/tex]

On substituting the known values,

[tex]\begin{gathered} Q=920\times79.7 \\ =73324\text{ Cal} \\ \approx73.3\text{ kCal} \end{gathered}[/tex]

Final answer:

The amount of heat that is to be removed to convert the given amount of water into ice is 73.3 kCal.

Therefore the correct answer is option C.

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