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Answer :
Answer:
4.68227 °C
Explanation:
[tex]m_o[/tex] = Mass of object = 500 kg
[tex]m_w[/tex] = 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
[tex]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[/tex]
The temperature change in the water is 4.68227 °C
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