a 70 kg hockey player moving at 8.0m/s stops quickly. how much ice melts due to the friction between the player's skates and the ice? assume that only 60% of the energy lost by the skater goes to melting the ice, and that the temperature of the ice is 0 degrees Celsius
To calculate the amount of ice that melts due to the friction between the player's skates and the ice, we need to find the energy lost by the skater and then determine how much of that energy is used to melt the ice.
The energy lost by the skater can be calculated using the principle of kinetic energy. The kinetic energy of the skater can be found using the formula:
KE = (1/2) * mass * velocity^2
Substituting the given values, we have:
KE = (1/2) * 70 kg * (8.0 m/s)^2
= 2240 J
Now, since only 60% of the energy lost by the skater goes to melting the ice, we need to find 60% of the calculated energy (2240 J):
Energy for melting ice = 0.60 * 2240 J
= 1344 J
To determine the amount of ice melted, we need to calculate the quantity of heat energy required to melt a given mass of ice. The specific heat of fusion for ice (amount of heat energy required to melt 1 kg of ice) is 334,000 J/kg.
Using this information, the mass of ice melted can be calculated using the formula:
Mass of ice melted = Energy for melting ice / Specific heat of fusion for ice
Substituting the values:
Mass of ice melted = 1344 J / 334,000 J/kg
= 0.004 kg
Therefore, approximately 0.004 kg (or 4 grams) of ice would melt due to the friction between the player's skates and the ice.
energy in stopping = 1/2 m v^2=1/2 *70*64=2240joules
energy in melting ice: 0.6* massice*333.6kj/kg
massice= 2.240*.6/333.6 kg = about 4 grams . check that
So if you had 1000 hockey players stopping on the ice, you would melt about 4 liters of ice. It wouldn't qualify as global warming.