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How much thermal energy must be removed from 3.75 kg of water to freeze it

at its melting point? Use Q = mass x latent heat of fusion

How much thermal energy must be removed from 3.75 kg of water to freeze it at its-example-1

1 Answer

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Answer: D.1,250kj

Explanation:

Hi, to answer this question we simply have to substitute mass= 3.75 and latent heat of fusion = 333 kj/kg in the equation given:

Q = mass x latent heat of fusion

Q = 3.75 kg x 333 kj/kg

Solving:

Q = 1,248 kj = 1,250kj (rounded)

The correct option is D.1, 250kj

Feel free to ask for more if needed or if you did not understand something.

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