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A laboratory determination of the specific latent heat of vaporisation of water uses a 120W heater to keep water boiling at its boiling point. Water is turned into steam at the rate of 0.050 g/s.Calculate the value of the specific latent heat of vaporisation obtained from this experiment. Show your working.
P= 120 WHeat (Q) = 120 J/smass = 0.05 g/s Q = mLL= 120/0.05= 2400 J/gthis should be right..
Guys what was the transformer formula? Np/Ns=Vp/Vs=Ip/Is?
see the thing here is, as u said they hav given the "mass" as 0.05g per second .. here i didnt literally mean it as 'mass' .. it was jus for ur understanding u kno..but did u get the concept ?? / sec cancels out in the equation ..
y-axis : 5 units = 0.25x-axis : 5 units = 0.25