# Question 7a804

Jun 17, 2016

$\text{2650 J}$

#### Explanation:

Let's assume that your textbook provided you with the molar enthalpy of fusion, $\Delta {H}_{\text{fus}}$, for water

$\Delta {H}_{\text{fus" = "6.01 kJ mol}}^{- 1}$

A substance's molar enthalpy of fusion tells you how much is needed in order for one mole of that substance to go from solid to liquid at its melting point.

color(purple)("solid")color(white)(a) "at melting point " stackrel(color(blue)(DeltaH_"fus")color(white)(aaa))(->) color(darkgreen)(" liquid") color(white)(a) "at melting point"

In your case, water's enthalpy of fusion tells you that you need $\text{6.01 kJ}$ of heat in order to convert one mole of ice, which is water in its solid form, to one mole of liquid water at ${0}^{\circ} \text{C}$.

The problem provides you with the mass of ice, so you must use water's molar mass to convert this to moles of ice

7.93 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "0.4402 moles H"_2"O"

So, if you need $\text{6.01 kJ}$ of heat to melt $1$ mole of ice at its melting point, it follows that $0.4402$ moles will require

0.4402 color(red)(cancel(color(black)("moles"))) * overbrace("6.01 kJ"/(1color(red)(cancel(color(black)("mole")))))^(color(blue)(= DeltaH_"vap")) = "2.6456 kJ"

Rounded to three sig figs and expressed in joules, $\text{1 kJ" = 10^3"J}$, the answer will be

"amount of heat needed" = color(green)(|bar(ul(color(white)(a/a)color(black)("2650 J")color(white)(a/a)|)))#