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

D.1,250kj

Step-by-step 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

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