Heat of Fusion of Ice

Mass A $-$ mass of calorimeter $108\mathrm{.}87$ g

Mass B $-$ mass of calorimeter and warm water $184\mathrm{.}57$ g

Mass C $-$ mass of calorimeter and water $($water $+$ ice melt$)191\mathrm{.}38$ g

Initial water temperature $($maximum$)34\mathrm{.}19\backslash$deg C

Final water temperature $($minimum$)24\mathrm{.}54\backslash$deg C $3)$ Calculate the heat lost by the cooling water using the equation Q $=$ delta T$$m$4\mathrm{.}186$ J$/$g$\backslash$deg C$,$ where Q $=$ heat $($in joules$),$ delta T $=$ change in temperature $($in $\backslash$deg C$)$ and m $=$ mass of water cooled $($in g$).$

$4)$ Determine the mass of ice melted $($mass of calorimeter and water $($original $+$ ice melt$)$ mass of calorimeter and warm water$).$

$5)$ Use your answers to Steps $3$ and $4$ to calculate the heat needed to melt $1$ g of ice $($J$/$g$).$

$6)$ An accepted value for the heat of fusion of ice is $334$ J$/$g$.$ Percentage difference is a standard way of measuring how far you are from an expected value. Calculate your percent difference using the formula:

$\%$ Difference $=\{[$calculated value $-$ accepted value$]/$accepted value$\}$ x $100$