Marc attaches a falling 250-kg object with a rope through a pulley to a paddle wheel shaft. He places the system in a well-insulated tank holding 12 kg of water. When the object falls, it causes the paddle wheel to rotate and churn the water. If the object falls a vertical distance of 100 m at constant speed, what is the temperature change of the water? (1 kcal = 4186 J, the specific heat of water is 4186 J/kg ·$\degree$C, and g = 9.8 m/s²) A) 19,600$\degree$C B) 4700$\degree$C C) 4.9$\degree$C D) 2.3$\degree$C

Question

Quizplus · Accepted Answer

The potential energy lost by the object is converted into heat energy, which increases the water's temperature. The potential energy lost is mgh = 250 kg * 9.8 m/s² * 100 m = 245,000 J. The heat energy absorbed by the water is Q = mcΔT, where m = 12 kg, c = 4186 J/kg°C. Solving for ΔT gives ΔT = Q/(mc) = 245,000 J / (12 kg * 4186 J/kg°C) ≈ 4.9°C.

Marc Attaches a Falling 250-Kg Object with a Rope Through °\degree°

Marc Attaches a Falling 250-Kg Object with a Rope Through $\degree$