Question

Calculate the change in entropy as 0.0419 kg of ice at 273.15 K melts. The latent heat of fusion of water is 333000 J/kg .

Answer 1

`\Delta S=51.1(J)/(K)`

Step-by-step explanation:

Hello!

In this case, since the entropy and enthalpy are related via the following equation for any process:

`\Delta S=(m\Delta H)/(T)`

As this problem relates the enthalpy and entropy of fusion, given the mass, melting temperature and heat of fusion, we compute the change in entropy of fusion as shown below:

`\Delta S=(0.0419kg*333,000(J)/(kg) )/(273.15K)\\\\\Delta S=51.1(J)/(K)`

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