It has been argued that skaters slide easily on ice because the pressure of the

Question

It has been argued that skaters slide easily on ice because the pressure of the skater’s blade creates a thin film of liquid water. The temperature of indoor ice rinks is typically at T = -7 C. What is the required mass in kg and lbm of the skater to melt the ice at this temperature? Does it seem reasonable that a thin film of water is created by the skater’s blade? You can use a blade thickness of 1.5 mm (~1/16 inch) and a length of 150 mm (~ 6 inches). Molar volume of liquid water (near 0 C) is 1.802 x 105 m3 mol1 and ice (near 0 C) is 1.965 x 105 m3 mol1. You can use a constant enthalpy of fusion (melting) of 6.01 kJ mol1. You can assume that the normal melting point of water is 0 C at 101.325 kPa. The slope of the solid-liquid saturation curve of water is negative for temperatures near the normal melting point.

Explanation / Answer

Here, required mass in kg and lbm of the skater = 6010 * 1.965 x 105 /(1.802 x 105 * 100)

                                                                             = 65.53 kg

                                                                             = 144.468 lbm

=>   Yes, it seem reasonable that a thin film of water is created by the skater’s blade .

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