1) Heat absorbed by water
[tex]q=mc\Delta T[/tex]
q= heat absorbed
m= mass (grams)
c=specific heat of water is 4.18 J/g-°C
ΔT= final temperature - initial temperature
2) Plug in known values
[tex]q=(330gH_2O)(4.18\frac{J}{gºC})(31ºC-27.4ºC)=4965.84J[/tex]
3) Heat transferred.
Heat absorbed by water is the heat released by the sample.
4) Heat transfer by sample
[tex]q=m\cdot\Delta Hf[/tex]
q= heat transfer
m= mass (grams)
ΔHf=enthalpy of fusion
5) Plug in known values and solve for ΔHf.
[tex]4965.84J=(115g)(\Delta Hf)[/tex][tex]\frac{4965.84J}{115g}=\frac{(115g)(\Delta Hf)}{115g}[/tex][tex]\frac{4965.84J}{115g}=\Delta Hf[/tex][tex]\Delta Hf=\frac{4965.84J}{115g}=43.18\frac{J}{g}[/tex]
The enthalpy of fusion of the sample is 43.18J/g.