A "clever" technician decides to heat some water for his coffee with an x-ray machine. If the machine produces 50 rad/s, how long will it take to raise the temperature of a cup of water by 55°C. Ignore heat losses during this time.
Find the speed in m/s of an alpha particle requires to come within 3.0E-14 m of a stationary gold nucleus. Then find the energy of the alpha particle in MeV.
Find the speed in m/s of an alpha particle requires to come within 3.0E-14 m of a stationary gold nucleus. Then find the energy of the alpha particle in MeV.
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According to my physics book, 1 rad is the amount of radiation that deposits 0.01 joules of energy into 1 kg of absorbing material.
50 rad/s = 0.5 joules/s per kilogram of absorbing material
http://www.bestfish.com/convert.html
According to the website above, 1 fluid ounce = 29.57 ml = 29.57 grams of water
1 cup = 8 fluid ounces
Mass of 1 cup of water = 8 * 29.57 = 236.56 grams = 0.23656 kg
Heat energy per second = 0.5 J/kg * 0.23656 kg = 0.11828 J
The x-rays supply 0.11828 Joules of energy to the water each second.
Heat energy required = mass of water * Specific heat of water * Temperature change
1 cup = 8 fluid ounces
Mass of 1 cup of water = 8 * 29.57 = 236.56 grams
Specific heat = 4.186 J/(g * ˚C)
Heat energy = 236.56 * 4.186 * 55 = 54,463.2088 Joules
This is the heat energy required to increase the temperature of 1 cup of water 55˚C.
This heat energy is supplied by the x-rays at the rate of 0.11828 Joules per second
0.11828 * time = 236.56 * 4.186 * 55
Time = (236.56 * 4.186 * 55) ÷ 0.11828
50 rad/s = 0.5 joules/s per kilogram of absorbing material
http://www.bestfish.com/convert.html
According to the website above, 1 fluid ounce = 29.57 ml = 29.57 grams of water
1 cup = 8 fluid ounces
Mass of 1 cup of water = 8 * 29.57 = 236.56 grams = 0.23656 kg
Heat energy per second = 0.5 J/kg * 0.23656 kg = 0.11828 J
The x-rays supply 0.11828 Joules of energy to the water each second.
Heat energy required = mass of water * Specific heat of water * Temperature change
1 cup = 8 fluid ounces
Mass of 1 cup of water = 8 * 29.57 = 236.56 grams
Specific heat = 4.186 J/(g * ˚C)
Heat energy = 236.56 * 4.186 * 55 = 54,463.2088 Joules
This is the heat energy required to increase the temperature of 1 cup of water 55˚C.
This heat energy is supplied by the x-rays at the rate of 0.11828 Joules per second
0.11828 * time = 236.56 * 4.186 * 55
Time = (236.56 * 4.186 * 55) ÷ 0.11828
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I have the answer to the 2nd question. If you make me one of your contacts and ask the 2nd question, I will sedn you the answer. Electron
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Potential electric energy = (k * Q * q)/r
Potential electric energy = (9 * 10^9 * 1.264 * 10^-17 * 3.2 * 10^-19) ÷ 3 * 10^-14
Kinetic energy of alpha particle = ½ * 6.64 * 10^-27 * v^2
set KE equal to PE and solve for velocity
Potential electric energy = (9 * 10^9 * 1.264 * 10^-17 * 3.2 * 10^-19) ÷ 3 * 10^-14
Kinetic energy of alpha particle = ½ * 6.64 * 10^-27 * v^2
set KE equal to PE and solve for velocity
Report Abuse
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Potential electric energy = (k * Q * q)/r
Potential electric energy = (9 * 10^9 * 1.264 * 10^-17 * 3.2 * 10^-19) ÷ 3 * 10^-14
Kinetic energy of alpha particle = ½ * 6.64 * 10^-27 * v^2
set KE equal to PE and solve for velocity
Potential electric energy = (9 * 10^9 * 1.264 * 10^-17 * 3.2 * 10^-19) ÷ 3 * 10^-14
Kinetic energy of alpha particle = ½ * 6.64 * 10^-27 * v^2
set KE equal to PE and solve for velocity
Report Abuse