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guajiro [1.7K]
3 years ago
13

Why is reaching activation energy necessary?

Physics
1 answer:
frez [133]3 years ago
5 0
Activation energy is the energy required for a reaction to occur, and determines its rate.<span>
</span>
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A large power plant heats 1917 kg of water per second to high-temperature steam to run its electrical generators.
erastova [34]

Complete Question

A large power plant heats 1917 kg of water per second to high-temperature steam to run its electrical generators.

(a) How much heat transfer is needed each second to raise the water temperature from 35.0°C to 100°C, boil it, and then raise the resulting steam from 100°C to 450°C? Specific heat of water is 4184 J/(kg · °C), the latent heat of vaporization of water is 2256 kJ/kg, and the specific heat of steam is 1520 J/(kg · °C).

J

(b) How much power is needed in megawatts? (Note: In real power plants, this process occurs under high pressure, which alters the boiling point. The results of this problem are only approximate.)

MW

Answer:

The heat transferred is  Q = 5.866 * 10^9 J

The power is  P = 5866\  MW

Explanation:

From the question we are told that

      Mass of the water per second is m = 1917 \ kg

      The initial temperature of the water is T_i  = 35^oC

      The boiling point of water is  T_b = 100^oC

      The final temperature T_f = 450^oC

      The latent heat of vapourization of water is  c__{L}} = 2256*10^3 J/kg

      The specific heat of water c_w = 4184 J/kg^oC

      The specific heat of stem is C_s =1520 \ J/kg ^oC

Generally the heat needed each second is mathematically represented as

         Q = m[c_w (T_i - T_b) + m* c__{L}}  + m* c__{S}} (T_f - T_b)]

Then substituting the value

        Q = m[c_w [T_i - T_b] + c__{L}}  + C__{S}} [T_f - T_b]]

         Q = 1917 [(4184) [100 - 35] + [2256 * 10^3]  +[1520]  [450 - 100]]

         Q = 1917 * [3.05996 * 10^6]

         Q = 5.866 * 10^9 J

The power required is mathematically represented as

         P = \frac{Q}{t}

From the question t = 1\ s

So  

        P = \frac{5.866 *10^9}{1}

        P = 5866*10^6 \ W

        P = 5866\  MW

6 0
3 years ago
What is the least amount of time required for a given point on this wave to move from y = 0 to y = 12cm?
ale4655 [162]

Time t = ?

<span>When wave is moving from y = 0  to y =12 cm</span>

By using the formula,

y = 15cos [(π/12) t)] = 0,

cos [(π/12) t)] = 0 = cos (π/2), so, 

(π/12)t = π/2,

t = (π/2) (12/π) 

t = 12/2

<span>t = 6 sec</span>

<span>so 6 sec is the least amount of time required</span>

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A body of mass 1.0 kg initially at rest slides down an incline plane that is 1.0 m high and 10.0 m long. If the body experiences
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Answer:

answer is

Explanation:

because

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Calculate the wavelength of a neutron that has a velocity of 200. cm/s. (the mass of a neutron = 1.675 × 10–27 kg and h = 6.63 ×
Licemer1 [7]
<span>To find the wavelength of a neutron can be calculated by using the formula Wavelength=h/m x v Where h is planck's constant m=mass of neutron v= velocity of the particle By substituting the given values Wavelength= 6.63 × 10–34 j s / 1.675 × 10–27 kg x 2 m/s^-1 Wavelength of a neutron=1.979 x 10^-7 m</span>
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Kg . Meter per second (Kg.m/s)
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