Question

Marc attaches a falling 500-kg object with a rope through a pulley to a paddle wheel shaft. He places the system in a well-insulated tank holding 25 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 = 4 186 J, the specific heat of water is 4 186 J/kgC, and g = 9.8 m/s2)

Answer 1

4.68227 °C

Step-by-step explanation:

`m_o` = Mass of object = 500 kg

`m_w` = Mass of water = 25 kg

c = Specific heat of water at 20°C = 4186 J/kg°C

h = Height from which the object falls = 100 m

g = Acceleration due to gravity = 9.8 m/s²

The potential energy and heat will balance each other

`PE=Q\\\Rightarrowmc m_ogh=m_oc\Delta T\\\Rightarrow \Delta T=(m_ogh)/(m_oc)\\\Rightarrow \Delta T=(500* 9.8* 100)/(25* 4186)\\\Rightarrow \Delta=4.68227\ ^(\circ)C`

The temperature change in the water is 4.68227 °C