Question

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?

Answer 1

Answer:

$m=33.9g$

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):

$Q=nCp \Delta T=3.00mol*75.38\frac{J}{mol\°C} *50.0\°C\\\\Q=11307J$

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

$Q=n \Delta _{fus}H$

So we solve for moles with the proper units handling:

$n=\frac{Q}{\Delta _{fus}H} =\frac{11307J}{6010\frac{J}{mol} } =1.88mol$

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

$m=1.88mol*\frac{18g}{1mol}\\ \\m=33.9g$

Best regards.