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sveticcg [70]
2 years ago
8

Suppose that you have left a 200-mL cup of coffee sitting until it has cooled to 30∘C , which you find totally unacceptable. You

r microwave oven draws 1100 W of electrical power when it is running. If it takes 45 s for this microwave oven to raise the temperature of the coffee to 60∘C , what is the efficiency of heating with this oven?
Physics
1 answer:
AleksAgata [21]2 years ago
8 0

Answer:

\eta=0.5074\ or\ 50.74\%

Explanation:

<em><u>Considering the density & specific heat capacity of coffee to be equal to that of water.</u></em>

<em><u>GIVEN:</u></em>

  • density \rho=1\ g.mL^{-1}
  • specific heat c=4.186\ J.g^{-1}.K^{-1}
  • mass of coffee, m=200\times 1=200\ g
  • initial temperature of coffee, T_i=30^{\circ}C
  • final temperature of coffee, T_f=60^{\circ}C
  • power rating of oven, P=1100\ W
  • time taken to reach the final temperature, t=35\ s

<u>Heat released by the coffee to come to 60°C:</u>

Q=m.c.\Delta T

Q=200\times 4.186\times 30

Q=[tex]\eta=\frac{25116}{49500}\ J[/tex]

<u>Now the energy used by the oven in the given time:</u>

E=P.t

E=1100\times 45

E=49500\ J

Now the efficiency:

\eta=\frac{Q}{E}

\eta=0.5074\ or\ 50.74\%

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Answer:

The gas was Hexane

Explanation:

taking the diference between the mass of the flask and the final mass qe can calculate the mass of liquid injected (assuming none escaped the flask):

m_{l}  = 27.4593g - 27.0928g = 0.3665g

with the volume of the flask we can get the density of the gas at the indicated pressure and temperature:

d_{g}  = \frac{0.3665 g}{0.1040L} = 3.524 g/L

From the ideal gases law we have that the density can be calculated as:

d_{g}  = \frac{P*M}{R*T}

Where R is the ideal gases constant = , and M the molecular weight of the fluid. Solving for M:

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2 years ago
A cell membrane consists of an inner and outer wall separated by a distance of approximately 10nm. Assume that the walls act lik
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Answer:

The options are approximations of the exact answers:

A) 1\times10^6N/C

B) 2\times10^{-13}N

C) 1\times10^{-2}V

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Explanation:

A) The electric field in a parallel plate capacitor is given by the formula E=\frac{\sigma}{k\epsilon_0}, where \epsilon_0=8.85\times10^{-12}C/Vm and in our case \sigma=10^5C/m^2 and, for air,k=1.00059, so we have:

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B) The K+ ion has one elemental charge excess, so its charge is q=1.6\times10^{-19}C, and the force a charge experiments under an electric field E is given by F=qE, so we have:

F=(1.6\times10^{-19}C)(1.13\times10^6N/C)=1.81\times10^{-13}N

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D) The electic field goes from positive to negative charges, so it goes towards the inner wall.

E) The work done by an electric field through a potential difference \Delta V on a charge Q is W=Q\Delta V, and is equal to the kinetic energy imparted on it, so we have:

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