bullyhunter77
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how would i do this?
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osv
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Set gravitational potential energy = specific heat capacity energy (assumes no thermal energy is transferred and also as the stream is continuous kinetic energy is likely constant so gpe goes straight to heat)
i.e mgh=mcΔθ and solve for θ
Energy gained by the water is not dependent on the mass as you can see the mass cancels from either side of the equation
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uberteknik
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(Original post by bullyhunter77)
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how would i do this?
The energy supplied to the water comes from the conversion of gravitational potential energy to heat.

The specific heat capacity of the water:

C = \frac{E}{m\Delta{T}}

where C = 4200Jkg-1K-1

m = mass

\Delta{T} is the temperature change

E = gravitational potential energy = mgh

and is the energy supplied to the water falling through a height of 100m

rearranging for \Delta{T}

\Delta{T} = \frac{E}{mC}

substituting for E = mgh

\Delta{T} = \frac{mgh}{mC} = \frac{gh}{C}

and plugging in the values:

\Delta{T} = \frac{9.81\text{ x }100}{4200} = 0.23K
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Alvie
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YokaiClan
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(Original post by bullyhunter77)
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how would i do this?
Basically E=mcΔθ
As no thermal energy, also E = mgh

So mgh = mcΔθ

gh = cΔθ

Δθ=gh / c

you know what height, specific heat capacity and gravitational field strength are, so u have enough information.
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