Evaporation of Water on a Jogger

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The discussion focuses on calculating the amount of sweat needed to cool a jogger's body temperature through evaporation. The latent heat of vaporization for water at body temperature is given as 2.38E+6 J/kg. The initial calculation for cooling the jogger by 1.10°C was incorrectly computed due to a misused specific heat capacity value. After correcting the specific heat capacity from 3350 to 3550 J/(kg*°C), the proper calculation yields a required evaporation of approximately 0.111 kg of sweat. The importance of accurate values in thermodynamic calculations is emphasized.
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The latent heat of vaporization of H2O at body temperature (37.0°C) is 2.38E+6 J/kg. To cool the body of a 71.9 kg jogger [average specific heat capacity = 3550 J/(kg*°C)] by 1.10°C, how many kilograms of water in the form of sweat have to be evaporated?

Q=cmt Q=mL

Q=3550(71.9)1.10=264951.5J
264951.5J=m(2.38E6) m=.111kg

the answer is wrong and i think I am pluggin in some wrong numbers somewhere. any help is appreciated
 
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I think you may have a calculator error, for Q=cmt, I get Q=(3550)(71.9)(1.10)=28769.5J. This may be your problem. See what you get as an answer using this value for Q.
 
yes that was it...i used 3350 instead of 3550 on accident. thank you.
 
No problem!
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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