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Idle Miggi

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Miggi

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Posted: 23 March 2012 at 7:53pm | IP Logged
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?

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Idle akhl

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akhl

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Posted: 24 March 2012 at 10:46pm | IP Logged
Originally posted by Miggi

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

Idle akhl

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akhl

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Posted: 24 March 2012 at 11:06pm | IP Logged
Originally posted by Miggi

A student wants to lose some weight. He knows that rigorous aerobic activity uses about 700 kcal/h (2900 kJ/h) and that it takes about 2000 kcal per day (8400 kJ) just to support necessary biological functions, including keeping the body warm. He decides to burn calories faster simply by sitting naked in a 16 C room and letting his body radiate calories away. His body has a surface area of about 1.7 m2 and his skin temperature is 35 C. Assuming an emissivity of 1.0, at what rate will this student "burn" calories? (kcal/h). If the student wants to lose 700kcal/day how much more time  will this take him per day, compares to rigorous aerobic activity?


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 ?



Idle akhl

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akhl

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Posted: 24 March 2012 at 11:20pm | IP Logged
Originally posted by Miggi

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 ?
 

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

Idle Miggi

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Miggi

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Posts: 142

Posted: 26 March 2012 at 9:38pm | IP Logged

@akhl yes it asks for how much more time it will take him to lose 700kca,/day, if its compares to aeroboic activity?

Would you please help me in the other questions as well? Thank u in advance :)

Originally posted by akhl

Originally posted by Miggi

A student wants to lose some weight. He knows that rigorous aerobic activity uses about 700 kcal/h (2900 kJ/h) and that it takes about 2000 kcal per day (8400 kJ) just to support necessary biological functions, including keeping the body warm. He decides to burn calories faster simply by sitting naked in a 16 C room and letting his body radiate calories away. His body has a surface area of about 1.7 m2 and his skin temperature is 35 C. Assuming an emissivity of 1.0, at what rate will this student "burn" calories? (kcal/h). If the student wants to lose 700kcal/day how much more time  will this take him per day, compares to rigorous aerobic activity?


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 ?



Idle akhl

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akhl

Joined: 30 March 2007

Posts: 5042

Posted: 27 March 2012 at 12:46pm | IP Logged
Originally posted by Miggi

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?


P1 = 1 atm
V1 = pr2 L = p * (4.00)^2 * 12.00 =  603.2 cm^3
T1 = 273 + 25 = 298 K

P1 V1 =n R T1
n = (P1 V1)/(R T1)
Substituting values,
n = 0.0246 mol

V2 = pr2 (L+x) = p * (4.00)^2 * (12.00 + 5.40) =  874.6 cm^3
deltaQ = 15.0 cal = 15.0 * 4.146 = 62.19 J
deltaQ = n c deltaT
deltaT = deltaQ/(n c) = 62.19/(0.0246 * 12.5) = 202 K
T2 = T1 + deltaT = 298 + 202 = 500 K

P1 V1/T1 = P2 V2/T2
P2 = P1 * (V1/V2) * (T2/T1)
P2 = 1 * (603.2/874.6) * (500/298)
P2 = 1.16 atm

Pressure outside = atmospheric pressure = 1 atm
Difference in pressure = 1.16 - 1 = 0.16 atm =  1.6 x 10^4 Pa
Force by spring = k x
Force due to difference in pressure = 1.6 x 10^4 x area
For equilibrium,

k x = 1.6 * 10^4 * area
k * 5.40 * 10^-2 = 1.6 * 10^4 * PI * (0.04)^2
k =  1489 N/m = 1500 N/m (rounded) (Answer)

Idle Miggi

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Miggi

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Posts: 142

Posted: 27 March 2012 at 6:53pm | IP Logged
if we see little or no DNA on gel, in which step of plasmids DNA isolation, electrophoresis, restriction digestion could the problem have occurred?

Idle akhl

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akhl

Joined: 30 March 2007

Posts: 5042

Posted: 27 March 2012 at 10:45pm | IP Logged
Originally posted by Miggi

True or False:The average velocity of the molecules of air in an enclosed room is zero even though the average speed is close to the speed of the sound.


True.
The molecules move randomly in different directions, making the average velocity zero.

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