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What mass of ice can be melted with the same quantity of heat as required to raise the temperature of 3.00 mol H2O(l) by 50.0°C?
[Afus H° = 6.01 kJ/mol H2O(s)]​

Respuesta :

sebassandin

Answer:

[tex]m=33.9g[/tex]

Explanation:

Hello,

In this case, we can first compute the heat required for such temperature increase, considering the molar heat capacity of water (75.38 J/mol°C):

[tex]Q=nCp \Delta T=3.00mol*75.38\frac{J}{mol\°C} *50.0\°C\\\\Q=11307J[/tex]

Afterwards, the mass of ice that can be melted is computed by:

[tex]Q=n \Delta _{fus}H[/tex]

So we solve for moles with the proper units handling:

[tex]n=\frac{Q}{\Delta _{fus}H} =\frac{11307J}{6010\frac{J}{mol} } =1.88mol[/tex]

Finally, with the molar mass of water we compute the mass:

[tex]m=1.88mol*\frac{18g}{1mol}\\ \\m=33.9g[/tex]

Best regards.

mass=33.9g

Given:

n= 3.00 mol

Δfus H° = 6.01 kJ/mol H₂O(s)

Enthalpy of fusion is the change in its enthalpy resulting from providing energy, typically heat, to a specific quantity of the substance to change its state from a solid to a liquid, at constant pressure.  

In this case, we can first compute the heat required for such temperature increase, considering the molar heat capacity of water (75.38 J/mol°C):

[tex]Q=nC_P[/tex]ΔT

[tex]Q=3.00\text{mol}*75.38\frac{J}{mol^oC} *50.0^oC\\\\Q=11307J[/tex]

Now, the mass of ice that can be melted is given by:

Q=nΔfus H°

So we solve for moles with the proper units handling:

n= Q/ Δfus H°

[tex]n=\frac{11307 J}{6010\frac{J}{mol} } =1.88 mol[/tex]

On substituting the moles with the molar mass of water we get:

[tex]m=1.88 mol*\frac{18 g}{1mol}\\\\m=33.9g[/tex]

The mass of ice is 33.9g.

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