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thanks akhl.. would u mind doing the other question as well??i am confused about it.

@Miggi,
Somehow the other question is not clear. Are you sure the questions are complete?
And are these two different questions?

@ akhl: there are two separate questions and from what i can understand is that the solution is first diluted to 2000ug/ml and then 2uL of that solution is taken.we need to convert it to ppm i guess? same for the 2nd question.

An experiment makes use of a water manometer attached to a flask. Initially the two columns in the gas manometer are at the same level and the air pressure in the flask and both sides of manometr is 1 atm. The experiment is set up when the air pressure is 7 degree Celcius. The left side of the manometer is connected to a flask and right side is capped so that the air at the end will be compressed when the flask is heated by a gas burner. The cap is initially 15cm above the water column. The volume of the flask is 1*10^4 m^3. When calculating the change in pressure assosiated with the heating of the gas in the flask, you can neglect the change in the volume of the gas(air in this cae) assosiated with the displacement of the water column in the manometer. Calculate how many calories (cal) have been added to the flask through heating from the gas burner given that specific heat of the air is 20.8 (J/K)/mol.

A 12.00cm cylindrical chamber has an 8.00cm diameter piston attached to one end. The piston is connected to an ideal spring. The piston is filled with helium gas. Initially the helium inside the chamber is at atmospheric pressure and at 25 degree celcius and the spring is not compressed. After 15.0 cal of heat is added to the temperature to raise its temperature to a new value, the spring is compressed by a distance of delta x = 5.40cm. If the specific heat of the helium is 12.5 J/K/mol, what is the spring constant of the spring? Edited by Miggi - 12 years ago

The thermal resistance of a seal's fur and blubber combined is 0.33 K/W. If the seal's internal temperature is 37C and the temperature of the sea is about 0C, what must be the heat output of the seal in order for it to maintain its internal temperature ? Edited by Miggi - 12 years ago

A rule of thumb states that time to hard boil an egg doublesfor every 10 degree celcius drop in temperature fromm 100 degree celcius. What activation energy approximately, does this rule imply for the chemical reactions that occur whenegg is being cooked?

Time doubles, which means speed of reaction becomes half of earlier.
k2 = k1/2
Or k1 = 2 k2

ln(k1/k2) = (Ea/R) (1/T2 - 1/T1)
ln(k1/k2) = (Ea/R) (T1 - T2)/(T1 T2)
ln(2) = (Ea/8.314) * (10)/ [(100+273)(100+273)]
ln(2) = Ea * 8.645 * 10^-6
Ea = 8.0 x 10^4 J/mol

Finding rate at hich calories burn
T = 35 + 273 = 308 K
To = 16 + 273 = 289 K
P = e * sigma * A * (T^4 - To^4)
P = 1.0 x 5.67 x 10^-8 x 1.7 x (308^4 - 289^4)
P = 195 W
P = 195/4.14 cal/s = 47.1 cal/s
P = 47.1 * 10^-3 * 3600 kcal/h
P = 170 kcal/h (Answer)

Finding time taken
For the second part, are you sure it asks how much more time? And is the value 700 kcal/day or 700 kcal/h ?

P = deltaT/R = (37-0)/0.33
= 112 W