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Llana [10]
3 years ago
9

How much energy is required to heat 70 g of water at 20°C to boiling

Physics
1 answer:
choli [55]3 years ago
3 0

Answer:

Q=23,430J

Explanation:

Hello,

In this case, since we compute the required energy via:

Q=mC\Delta T

Whereas m is the mass which here is 70 g, C the specific heat which for water is 4.184 J/(g°C) and ΔT is the temperature difference which is:

\Delta T=100-20=80\°C

Therefore, the energy turns out:

Q=70g*4.184\frac{J}{g\°C}*80\°C\\ \\Q=23,430J

Best regards.

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Assume that the home construction industry is perfectly competitive and in long-run competitive equilibrium. It follows that: A.
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(a) How many fringes appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit
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Answer:

a

The number of fringe is  z  = 3 fringes

b

The  ratio is I = 0.2545I_o

Explanation:

a

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        The distance between the slit is  d = 0.117mm = 0.117 *10^{-3} m

        The width of the slit is  a = 35.7 \mu m = 35.7 *10^{-6}m

let  z be the number of fringes that appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit pattern is  and this mathematically represented as

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Substituting values

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b

   From the question  we are told that the order  of the bright fringe is  n = 3

   Generally the intensity of  a pattern  is mathematically represented as

                 I = I_o cos^2 [\frac{\pi d sin \theta}{\lambda} ][\frac{sin (\pi a sin \frac{\theta}{\lambda } )}{\pi a sin \frac{\theta}{\lambda} } ]

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 And  Generally  sin \theta = \frac{n \lambda }{d}

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               I = I_o cos^2 (n \pi)[\frac{\frac{sin(\pi a (\frac{n \lambda}{d} ))}{\lambda} )}{ \frac{ \pi a (\frac{n \lambda }{d} )}{\lambda} } ]

               I = I_o cos^2 (3 \pi) [\frac{sin (\frac{3 \pi }{6} )}{\frac{3 \pi}{6} } ]

                I = I_o (1)(0.2545)

                  I = 0.2545I_o

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