Melting under a skate? It has been postulated that the low friction between a sk
ID: 1056578 • Letter: M
Question
Melting under a skate? It has been postulated that the low friction between a skate and ice is due to melting of the ice under the skate. You are to estimate whether this is true or not. For this purpose you must have the Gibbs energy of ice and water as a function of pressure and temperature. The effect of pressure on both is small, but the pressure under a skate is quite high. The pressure effect stems from the pressure term in the enthalpy. We take the reference temperature and pressure equal to o degree C and 0.1 MPa. Near this point: g^L = v^L(p-p_ref) g^s = delta h^melt(T/T_ref -l)+v^s'(p-p_ref) The specific volumes of water and ice are: v^L =l0^-3m^3kg^-1 v^s =1.09xl0^-3 m^3 kg^1 The enthalpy of melting is delta h^melt = 334 kJ kg^-1 Estimate the pressure under a skate for a person with a weight of 8oo N (i.e. 8o kg). The skate is 30 cm long and 1 mm wide. What will the melting temperature be at this pressure?Explanation / Answer
we knwo that
weight = mg = 800 N
where m = mass
g = acceleration due to gravity = 9.8 m/s^2
Now pressure can be calculated as
Pressure = Force exerted /Area = mass X acceleration due to gravity / lenght X bredth
= (80)(9.8)/(30 x 10^-2)(1x10^-3) = 26.13 X 10^5 N / m^2
We know that the relation between atmospere (unit of pressure) and N / m^2 is
1 atmosphere = 101,325 N/m^2
therefore
26.13 X 10^5 N / m^2 = 26.13 X 10^5 / 101,325 atm = 25.788 atm
This pressure will be exerted by person
The atmospheric pressure = 1atm
so total pressure = 26.788 atmosphere
we know that
lnP2/P1 = -Delta Hvap / R (1/T2 -1/T1)
ln (26.788 / 1) = -40.66 / 8.314 X 10^-3 (1/T2 - 1/T1)
-3.288 / 4890.55 = (1/T2 - 1/273)
-6.723 / 10000 = -0.0006723 + 0.00366 = 1/T2
T2 = 1/0.00299 = 334.45 Kelvin