A 6 kg object falls 10 m. The object is attached mechanically to a paddle-wheel which rotates as the object falls. The paddle-wheel is immersed in 600 g of water at 15°C. What is the maximum temperature that the water could be increased to? Is this a very efficient way of heating water? The specific heat of water is 4.186 J/(g* ˚C).

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nuuk nuuk · Answer 1 · 2019-09-19T10:32:22+02:00

Answer:[tex]16.234^{\circ}C[/tex]

Explanation:

Given

Mass of object (m)=6 kg

falling height(h)=10 m

mass of water([tex]m_w[/tex])=600 gm

temperature of water =15

specific heat of water [tex]=4.186 j/g-^{\circ}C[/tex]

Let T be the Final Temperature of water

Here Object Potential Energy is converted into Heat energy which will be absorbed by water

Potential Energy(P.E.)[tex]=mgh=6\times 9.81\times 10=588.6 J[/tex]

Heat supplied[tex]=m_wc(\Delta T)[/tex]

[tex]H.E.=600\times 4.186\times (T-16) [/tex]

[tex]588.6=2511.6\times (T-16)[/tex]

T-16=0.234

[tex]T=16.234^{\circ}C[/tex]

This is not an efficient way of heating water as there is only[tex] 0.234^{\circ}C [/tex]increase in temperature.