Marc attaches a falling 500-kg object with a rope through apulley to a paddle wheel shaft. He places the system in a well-insulated tank holding 25 kg of water. When the object falls, itcauses the paddle wheel to rotate and churn the water. If theobject falls a vertical distance of 100 m at constant speed, what isthe temperature change of the water? (1 kcal = 4 186 J, thespecific heat of water is 4 186 J/kg C, and g = 9.8 m/s2).A-19600℃B-4700℃C-4.7℃D-0.8℃
Question
Marc attaches a falling 500-kg object with a rope through apulley to a paddle wheel shaft. He places the system in a well-insulated tank holding 25 kg of water. When the object falls, itcauses the paddle wheel to rotate and churn the water. If theobject falls a vertical distance of 100 m at constant speed, what isthe temperature change of the water? (1 kcal = 4 186 J, thespecific heat of water is 4 186 J/kg C, and g = 9.8 m/s2).A-19600℃B-4700℃C-4.7℃D-0.8℃
Solution
First, we need to calculate the potential energy of the falling object. The formula for potential energy is PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height.
PE = 500 kg * 9.8 m/s^2 * 100 m = 490,000 J
This energy is transferred to the water, causing it to heat up. The amount of heat energy required to raise the temperature of a substance is given by the formula Q = mcΔT, where m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
We can rearrange this formula to solve for ΔT: ΔT = Q / (mc)
ΔT = 490,000 J / (25 kg * 4,186 J/kg°C) = 4.7°C
So, the temperature change of the water is 4.7°C. The correct answer is C-4.7℃.
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