Isoproply alchohol C3H7OH can be used as a cooling run for hospital patients. How much heat in (kj) is dissipated from the skin of a person who is rubbed with 3.0 mL of isoproply alcohol?
I presume this is using enthalpy of fusion, so if I am incorrect feel free to correct me if you haven’t covered this yet.
In this case, we need some thermochemistry data about the isopropanol (which is another name for isopropyl alcohol), and here’s what I was able to dig up from the NIST web book and Chemspider:
molecular mass = 60.095 g/mole
So, we know the volume and the density of the isopropanol, so we can get the mass easily knowing the definition of denisty:
So, Mass= Density*Volume
Or in with values:
Mass = 0.785 g/mL * 3.0 mL =2.355 grams of isopropanol
Now, dividing by the molar volume will give us the number of moles, which we can use with the enthalpy of vaporization:
Moles = mass/molecular mass
Moles = 2.355 g/ 60.095 g/mole
Moles = 3.91879524086x10^-2 moles
Now, we know how many moles of isopropanol we have, as well as its heat of vaporization, all we have to do now is multiply the two together!
Heat dissipated= dH(fus)*moles of isopropanol
Heat dissipated= 45 kJ/mole * 3.91879524086x10^-2 moles
Heat dissipated=1.763457858 kJ or 1763.457858 J
Rounding to 2 sig figs we have either 1.8 kJ or 1.8*10^3 J of heat dissipated!