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Dominik [7]
3 years ago
6

How much heat is required to convert 500g of liquid water at 28°C into steam at 150 °C? Take the specific heat capacity of water

to be 4183 J/Kg °C and the latent heat of vaporization to be 2.26 × 10^6 J/Kg.
Physics
1 answer:
storchak [24]3 years ago
6 0

Answer:

Q = 1.404 × 10^(5) KJ

Explanation:

We are given:

Mass;m = 500 g = 0.5kg

Temperature 1;T1 = 28 °C

Temperature:T2 = 150 °C

Specific heat capacity;c_p = 4183 J/Kg °C

Latent heat of vaporization;L = 2.26 × 10^(6) J/Kg.

The heat energy needed is given by;

Q = sensible heat energy + Latent heat

Formula for sensible heat is;

Sensible heat energy = mc(t2 - t1)

Formula for Latent heat is ;

Latent heat = mL

Thus:

Q = mc(t2 - t1) + mL

Q = m[c(t2 - t1) + L]

Q = 0.5((4183(159 - 28) + (2.26 × 10^(6)))

Q = 1.404 × 10^(8) J = 1.404 × 10^(5) KJ

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Two vectors are given by a = 8.6i + 5.1 j and b = 931 + 9.5.
skelet666 [1.2K]

Answer:

<u>Assuming b = 9.3i + 9.5j</u> <em>(b = 931 + 9.5 is wrong):</em>

a) a×b = 34.27k

b) a·b = 128.43

c) (a + b)·b = 305.17

d) The component of a along the direction of b = 9.66

Explanation:

<u>Assuming b = 9.3i + 9.5j</u> <em>(b = 931 + 9.5 is wrong)</em> we can proceed as follows:  

a) The vectorial product, a×b  is:

a \times b = (8.6*9.5 - 5.1*9.3)k = 34.27k

b) The escalar product a·b is:

a\cdot b = (8.6*9.3) + (5.1*9.5) = 128.43

c) <u>Asumming (a</u><u> </u><u>+ b)·b</u> <em>instead a+b·b</em> we have:

(a + b)\cdot b = [(8.6 + 9.3)i + (5.1 + 9.5)j]\cdot (9.3i + 9.5j) = (17.9i + 14.6j)\cdot (9.3i + 9.5j) = 305.17

d) The component of a along the direction of b is:

a*cos(\theta) = \frac{a\cdot b}{|b|} = \frac{128.43}{\sqrt{9.3^{2} + 9.5^{2}}} = 9.66

I hope it helps you!                        

5 0
3 years ago
Once an object enters orbit, what keeps the object moving sideways?
zubka84 [21]

Answer: Inertia!!

Explanation: I just completed the edg quiz and got that answer correct! Hope its not too late for you!

7 0
3 years ago
If the wavelength is changed to λ/2, does the central spot remain bright, does the central spot become dark, or do the fringes d
omeli [17]

Answer:The central spot becomes Dark

Explanation: it become dark because as the wavelength reduces,the velocity in the detector decreases, this time by 90degrees

4 0
3 years ago
A diver jumps off a diving platform that is 20 meters long. Describe the transfer of energy that occurs during the fall.
kobusy [5.1K]

Answer: gravitational potential energy is converted into kinetic energy

Explanation:

When the diver stands on the platform, at 20 m above the surface of the water, he has some gravitational potential energy, which is given by

E=mgh

where m is the man's mass, g is the gravitational acceleration and h is the height above the water. As he jumps, the gravitational potential energy starts decreasing, because its height h above the water decreases, and he acquires kinetic energy, which is given by

K=\frac{1}{2}mv^2

where v is the speed of the diver, which is increasing. When he touches the water, all the initial gravitational potential energy has been converted into kinetic energy.

8 0
3 years ago
Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e.
alexandr1967 [171]

Answer:

v=\sqrt{k\frac{e^2}{m_e r}}, 2.18\cdot 10^6 m/s

Explanation:

The magnitude of the electromagnetic force between the electron and the proton in the nucleus is equal to the centripetal force:

k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}

where

k is the Coulomb constant

e is the magnitude of the charge of the electron

e is the magnitude of the charge of the proton in the nucleus

r is the distance between the electron and the nucleus

v is the speed of the electron

m_e is the mass of the electron

Solving for v, we find

v=\sqrt{k\frac{e^2}{m_e r}}

Inside an atom of hydrogen, the distance between the electron and the nucleus is approximately

r=5.3\cdot 10^{-11}m

while the electron mass is

m_e = 9.11\cdot 10^{-31}kg

and the charge is

e=1.6\cdot 10^{-19} C

Substituting into the formula, we find

v=\sqrt{(9\cdot 10^9 m/s) \frac{(1.6\cdot 10^{-19} C)^2}{(9.11\cdot 10^{-31} kg)(5.3\cdot 10^{-11} m)}}=2.18\cdot 10^6 m/s

7 0
3 years ago
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