Marc attaches a falling 328-kg object with a rope through a pulley to a paddle wheel shaft.He places the system in a well-insulated tank holding 23.2 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 88.6 m at constant speed,what is the temperature change of the water? (1 kcal = 4 186 J,the specific heat of water is 4 186 J/kg⋅°C,and g = 9.80 m/s²) A)17 000 C° B)2 930 C° C)2.93 C° D)0.587 C° E)5.87 C°

Question

Quizplus · Accepted Answer

First, we need to find the gravitational potential energy that is converted into thermal energy in the water:
ΔE = mgh = (328 kg) x (9.80 m/s^2) x (88.6 m) = 286078.4 J

Next, we need to find the mass of the water that is heated:
m_water = 23.2 kg

Then, we use the specific heat of water to find the change in temperature:
ΔT = ΔE / (m_water x c_water) = 286078.4 J / (23.2 kg x 4,186 J/kg⋅°C) ≈ 2.93°C

Therefore, the answer is C) 2.93°C.

Marc Attaches a Falling 328-Kg Object with a Rope Through