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Dmitrij [34]
3 years ago
5

They want a mixture of 150 grams of water at 60 degrees Celsius, using hot water at 90 degrees Celsius and cold water at 14 degr

ees Celsius, how many grams of water should be used per face?​
Physics
1 answer:
Gnesinka [82]3 years ago
6 0

Answer:

Explanation:

Here,

Final mass(m)=150g

Final Temperature(t)=60°C

Temperature of hot water(t1)=90C

Temperature of cold water(t2)=14C

Let the grams of water to be used per face be m1 grams and (150-m1) grams for hot and cold water respectively.

We know that,

Heat lost by hot body=Heat gained by cold body

or, m1 s1 Δt1=m2 s2 Δt2

or, m1 (t1-t)=(150-m1) (t-t2)  [∵s1=s2, SHC of water is taken to be same]

or, m1 (90-60)=(150-m1) (60-14)

or, 30m1=6900-46m1

or, 76m1=6900

∴m1=90.79g

(150-m1)=59.21g

Hence, 90.79g and 59.21 g hot and cold water must be used respectively.

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The first law of Thermodynamics is another way to describe the law of conservation of Energy. It states that:
nikitadnepr [17]

Answer:

C. The change of internal energy of a system is the sum of work and heat spent on it.

Explanation:

The law of conservation of Energy states that energy cannot be destroyed but can only be converted or transformed from one form to another. Therefore, the sum of the initial kinetic energy and potential energy is equal to the sum of the final kinetic energy and potential energy.

Mathematically, it is given by the formula;

Ki + Ui = Kf + Uf .......equation 1

Where;

Ki and Kf are the initial and final kinetic energy respectively.

Ui and Uf are the initial and final potential energy respectively.

The law of conservation of Energy is another way to describe the law of Thermodynamics. It states that the change of internal energy of a system is the sum of work and heat spent on it.

Mathematically, it is given by the formula;

ΔU = Q − W

Where;

ΔU represents the change in internal energy of a system.

Q represents the net heat transfer in and out of the system.

W represents the sum of work (net work) done on or by the system.

6 0
3 years ago
Imagine that Earth stops orbiting the Sun but continues to rotate in place about its own axis at its current rate. In this case,
otez555 [7]

Answer:

The length of the solar day will get shorter.

Explanation:

  • The blue planet Earth not only rotates around it's own axis but also rotates around the Sun and everyday it moves a little bit around the axis.
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7 0
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A boy is exerting a force of 70 N at a 50-degree angle on a lawn mower. He is accelerating at 1.8 m/s2. Round the answers to the
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are you in middle school

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8 0
3 years ago
How do the dark lines of an atom''s absorption spectrum relate to the bright lines of its emission spectrum?
tangare [24]

Wouldn't it be neat if an electron falling closer to the nucleus ... emitting a
photon ... actually gave out more energy than it needed to climb to its original
energy level by absorbing a photon !   If there were some miraculous substance
that could do that, we'd have it made.

All we'd need is a pile of it in our basement, with a bright light bulb over the pile,
connected to a tiny hand-crank generator.

Whenever we wanted some energy, like for cooking or heating the house, we'd
switch the light bulb on, point it towards the pile, and give the little generator a
little shove.  It wouldn't take much to git 'er going.

The atoms in the pile would absorb some photons, raising their electrons to higher
energy levels.  Then the electrons would fall back down to lower energy levels,
releasing more energy than they needed to climb up.  We could take that energy,
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the house or run the dishwasher.

The energy an electron absorbs when it climbs to a higher energy level (forming
the atom's absorption spectrum) is precisely identical to the energy it emits when
it falls back to its original level (creating the atom's emission spectrum).

Energy that wasn't either there in the atom to begin with or else pumped
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You get what you pay for, or, as my grandfather used to say, "For nothing
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3 0
3 years ago
Use the Bohr model to address this question. When a hydrogen atom makes a transition from the 5 th energy level to the 2nd, coun
iris [78.8K]

Answer:

A. 2.82 eV

B. 439nm

C. 59.5 angstroms

Explanation:

A. To calculate the energy of the photon emitted you use the following formula:

E_{n1,n2}=-13.4(\frac{1}{n_2^2}-\frac{1}{n_1^2})     (1)

n1: final state = 5

n2: initial state = 2

Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):

E_{5,2}=-13.6(\frac{1}{5^2}-\frac{1}{2^2})=2.82eV

B. The energy of the emitted photon is given by the following formula:

E=h\frac{c}{\lambda}   (2)

h: Planck's constant = 6.62*10^{-34} kgm^2/s

c: speed of light = 3*10^8 m/s

λ: wavelength of the photon

You first convert the energy from eV to J:

2.82eV*\frac{1J}{6.242*10^{18}eV}=4.517*10^{-19}J

Next, you use the equation (2) and solve for λ:

\lambda=\frac{hc}{E}=\frac{(6.62*10^{-34} kg m^2/s)(3*10^8m/s)}{4.517*10^{-19}J}=4.39*10^{-7}m=439*10^{-9}m=439nm

C. The radius of the orbit is given by:

r_n=n^2a_o   (3)

where ao is the Bohr's radius = 2.380 Angstroms

You use the equation (3) with n=5:

r_5=5^2(2.380)=59.5

hence, the radius of the atom in its 5-th state is 59.5 anstrongs

8 0
3 years ago
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