In an isothermal expansion at 27 degree C, an ideal as does 60J of work. W hat i
ID: 1997527 • Letter: I
Question
In an isothermal expansion at 27 degree C, an ideal as does 60J of work. W hat is the change in entropy of the gas? An ideal gas is under an initial pressure of 2.45 times 10^4 Pa and occupies a volume of 0.20 m^3. The slow addition of 8.4 times 10^3 J of heat to this gas causes it to expand isobarically to a volume of 0.40 m^3. How much work is done by the gas in the process. When resting, a person gives off heat at a rate of about 100W. If the person is submerged in a tub containing 150kg of water 27 degree C and the heat from the person goes only into the water, how many hours will it take for the water temperature to rise to 28 degree C.Explanation / Answer
1. temperature T = 27 0C = 300 k, woek done W = 60 J, change in entropy dS=?
from the relation dW = T*dS ==> dS = dW/T = 60 /300 = 0.2 J/k
2.
pressure P1 = 2.45*10^4 Pa, volume expanded from v1 = 0.2 m^2, to V2 = 0.40 m^3
work done W = 8.4*10^3 J,
work done in isobaric process W P*dV = 2.45*10^4(0.4-0.2) J = 4900 J
3.
rate of heat from a person gives off is 100 W,
if the person submerged in a tub of water of mass m = 150 kg, oat temperature T = 27 0C = 300 k
if the heat only goes to water from person then the time taken to raise the temeperature of water from 27 0C to 28 0C is
Q/t = m*c*(T2-T1)
t = Q/m*c*(T2-T1)
= 100/(150*4180(1)) s
t = 0.0001594896332 s
t = 159.49 micro seconds
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