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Paladinen [302]
3 years ago
14

The specific heat of fat is roughly 1,700 J/(kgoC) whereas that for water is around 4200 J/(kgoC). A 0.63 kg solution of lipids

and water is heated at a rate of 100 W for 5 minutes. During this time, the temperature of the system increases by 20oC. What is the mass of the lipid content, to the nearest hundredth of a kg, in this solution?
Physics
1 answer:
BartSMP [9]3 years ago
3 0

Answer:

the mass of the lipid content, to the nearest hundredth of a kg, in this solution =0.46 kg

Explanation:

Total heat content of the fat = heat content of water +heat content of the lipids

Let it be Q

the Q= (mcΔT)_lipids + (mcΔT)_water

total mass of fat  M= 0.63 Kg

Q= heat supplied = 100 W in 5 minutes

ΔT= 20°C

c_lipid= 1700J/(kgoC)

c_water= 4200J/(kgoC)

then,

100\times5\times60= m(1700)20+(0.63-m)(4200)20

solving the above equation we get

m= 0.46 kg

the mass of the lipid content, to the nearest hundredth of a kg, in this solution =0.46 kg

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The first ionization energy of a hydrogen atom is 2.18 aj (attojoules). what is the frequency and wavelength, in nanometers, of
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1) Frequency: 3.29\cdot 10^{15}Hz

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E=hf

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f=\frac{E}{h}=\frac{2.18\cdot 10^{-18} J}{6.63\cdot 10^{-34} Js}=3.29\cdot 10^{15} Hz


2) Wavelength: 91.2 nm

The wavelength of the photon can be found from its frequency, by using the following relationship:

\lambda=\frac{c}{f}

where c=3\cdot 10^8 m/s is the speed of light and f is the frequency. Substituting the frequency, we find

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3 years ago
A burglar attempts to drag a 108 kg metal safe across a polished wood floor Assume that the coefficient of static friction is 0.
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Answer:

2.00 m/s²

Explanation:

Given

The Mass of the metal safe, M = 108kg

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Co-efficient of kinetic friction, \mu_k = 0.3

Now,

The force against the kinetic friction is given as:

f = \mu_k N = u_k Mg

Where,

N = Normal reaction

g= acceleration due to the gravity

Substituting the values in the above equation, we get

f = 0.3\times108\times9.8

or

f = 317.52N

Now, the net force on to the metal safe is

F_{Net}= F-f

Substituting the values in the equation we get

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or

F_{Net}= 216.48

also,

 

F_{Net}= M\timesacceleration of the safe

Therefore, the acceleration of the metal safe will be

acceleration of the safe=\frac{F_{Net}}{M}

or

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or

 

acceleration of the safe=2.00 m/s^2

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3 years ago
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