In: Chemistry
The following in a thermodynamics question. I believe the change in Gibbs Free Energy needed for ice melting is 333 J/g. Please show all steps/calculations in your solution. Thanks!
It is often said that pressure-induced melting is the reason ice skating is possible. Calculate how heavy a person would need to be to melt ice under his or her skates at T = -10 C. The blade edge of a typical skate is about one-eighth of an inch wide and about 12 inches long. Give your answer in pounds. Furthermore, according to this notion of pressure-induced melting as the basis for ice skating, what is the lowest temperature at which 145-lb. Apolo Anton Ohno could perform? Repeat your calculation for a leaden sculpture of Apolo. (To figure out the last part, you will need Apolo's density. For this purpose, assume that he is composed entirely of liquid water.)
∆Hfusion = 333 J/g
Heat needed to melt ice=Q=mC∆T + m ∆Hfusion
For m=1g of ice
C=heat capacity of ice=2.03 J/g K
Heat needed to melt ice=Q=m C∆T + m ∆Hfusion=1g*2.03 J/g K*(273K-263K)+1g*333J/g=353.3 J/g
Now pressure applied* change of volume of Ice as it becomes water=P*Volume=Q
P=m*g/Area under skates [g=acceleration due to gravity]
=mass of person(M) *9.8m/s^2/(1/8”*12”)
=9.8 M* ms^-2/1.5 sq inch
1 inch=0.0254 m
1 in^2=(0.0254 m)^2=0.000645 m^2
P=9.8M ms^-2/(1.5 sq inch *0.000645 m^2/in^2)=10123.97 M ms^-2/m^2
Volume of ice under skates=V=mass/density of ice=1g/0.917 g/cm3
[density is taken at 0 degree or273K]
V=1.09 cm3/g
So P*V=10123.97 M s^-2/m * 1.09 cm3/g *10^-6 m3/cm3=Q=353.3 J/g
Or 10123.97 M*1.09*10^-6 =353.3
M=0.032*10^6g=0.032*10^6*10^-3 kg=0.032*10^3kg=32 kg weight(answer)
1lb=0.453kg
So weight of Apolo Anton Ohno=145*0.453kg=65.68 kg
P*V=mg*1.09*10^-6 m3=65.68 kg*9.8 ms^-2*1.09*10^-6 m3
Q= m C∆T + m ∆Hfusion=1g*2.03 J/g K*∆T+1g*333J/g
P*V=Q
Or,2.03J/K *∆T+333J=65.68 kg*9.8 ms^-2*1.09*10^-6 m3=701.65 *10^-6 J
Or, Or,2.03J/K *∆T+333J=701.65 *10^-6 J=0.000701J
Or, 2.03J/K *∆T=-332.9 J
∆T=-164.03K=273-T
T=-164.03-273=-437K=-164.03C