Heat of Fusion of Ice

Mass A $-$ mass of calorimeter $12\mathrm{.}89$ g

Mass B $-$ mass of calorimeter and warm water $89\mathrm{.}82$ g

$*$Note: Start data collection after stabilizing water temp before weighing and adding dry ice to minimize sublimation losses that will affect your experiment results.

Mass C $-$ mass of dry ice before adding to water $($use ~$5$g$)8\mathrm{.}73$ g

Initial water temperature $($maximum$)36\mathrm{.}02\backslash$deg C

Final water temperature $($minimum$)17\mathrm{.}55\backslash$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$).$

$4)$ Use your answers to Steps $3$ and the mass of dry ice before it was added to the water to calculate the heat needed to sublime $1$ g of dry ice $($J$/$g$).$

$5)$ An accepted value for the heat of sublimation of dry ice is $571$ 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 below:

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