Marc attaches a falling 400-kg object with a rope through a pulley to a paddle wheel shaft.He places the system in a well-insulated tank holding 30 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 120 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 = 9.8 m/s²)

Question

Quizplus · Accepted Answer

The potential energy lost by the falling object is equal to the work done by the paddle wheel on the water. 
Potential energy lost by the object = mgh = 400 kg * 9.8 m/s² * 120 m = 470,400 J 
Work done on the water = 470,400 J 
We can use the formula for specific heat: 
Q = mcΔT 
Where Q is the amount of heat transferred, m is the mass of the water, c is the specific heat of water, and ΔT is the temperature change. 
We know that Q = 470,400 J, m = 30 kg, and c = 4,186 J/(kg·°C). Solving for ΔT: 
ΔT = Q / (mc) = 470,400 J / (30 kg * 4,186 J/(kg·°C)) = 3.7°C

Therefore, the answer is C) 3.7 C°.

Marc Attaches a Falling 400-Kg Object with a Rope Through