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

Answer:

4.68227 °C

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

$PE=Q\\\Rightarrowmc m_ogh=m_oc\Delta T\\\Rightarrow \Delta T=\frac{m_ogh}{m_oc}\\\Rightarrow \Delta T=\frac{500\times 9.8\times 100}{25\times 4186}\\\Rightarrow \Delta=4.68227\ ^{\circ}C$

The temperature change in the water is 4.68227 °C