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Water is boiling in a kettle on an electric hot-plate of 800 W power. Find the steam outflow velocity, if the cross section of the spout is 0.9 cm and the pressure at the output is normal. The officiency of the hot plate is 72%. |
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Answer» Solution :The outflow VELOCITY is `v=mu//rho S`, where `mu` is the amount of water that evaporates per SECOND. Obviously, `mu=(Q)/(Lt)=(etaP)/(L),` WHEREP is the power of the hot-plate and `eta` is its efficiency. HENCE `v=etaP// rhoLS` |
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