What Is The Temperature Difference In Water At The Top And Bottom Of A Series Of Waterfalls?
If traveling at 8 m/s at top and 3 m/s at bottom. Overall height of waterfall 15m
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The initial kinetic and potential energy is presumed to be converted to thermal energy and the final kinetic energy.
((mv12/2+mgh1) = (mv02+mgh0)) + m(specific heat of H2O)*(temperature rise)
We will use T for the temperature rise.
kg((8m/s)2/2 + (9.8m/s2)(15 m)) = kg((3m/s)2/2 + 0) + kg(4.186 kgm2/s2/g/K)T
(Note that kg/g = 1000.)
Dividing out the mass (kg), we get
(64/2 + 9.8*15) m2/s2 = 9/2 m2/s2 + (4186m2/s2/K)T
Subtract 4.5 m2/s2 from both sides, and divide by 4186m2/(s2K)
(((32+147)-4.5)/4186)K = T = .042K rise in temperature
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This is an interesting question. It hadn't really occurred to me that water might get warmer just by slowing down. I suppose the small rise in temperature is not something that our senses would register, so it is not within our direct experience.
((mv12/2+mgh1) = (mv02+mgh0)) + m(specific heat of H2O)*(temperature rise)
We will use T for the temperature rise.
kg((8m/s)2/2 + (9.8m/s2)(15 m)) = kg((3m/s)2/2 + 0) + kg(4.186 kgm2/s2/g/K)T
(Note that kg/g = 1000.)
Dividing out the mass (kg), we get
(64/2 + 9.8*15) m2/s2 = 9/2 m2/s2 + (4186m2/s2/K)T
Subtract 4.5 m2/s2 from both sides, and divide by 4186m2/(s2K)
(((32+147)-4.5)/4186)K = T = .042K rise in temperature
_____
This is an interesting question. It hadn't really occurred to me that water might get warmer just by slowing down. I suppose the small rise in temperature is not something that our senses would register, so it is not within our direct experience.
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Excellent, thanks. Should your answer be .042 C instead of K?..........edit.. Sorry my mistake, you are spot on. Good work
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