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Marianna [84]
3 years ago
13

1.3 kg of water, at an initial temperature of 25oC, is heated at a rate of 100 W in a well-insulated container. How much time (i

n minutes) will it take for the water to reach the boiling point (100oC)
Physics
1 answer:
Leviafan [203]3 years ago
6 0

Answer:

68.25 minutes.

Explanation:

Power = Energy/time or

Power = Heat/time

P = Q/t....................... Equation 1

Q = Pt ........................ Equation 2

Where Q = quantity of heat, P = power, t = time.

Also,

Q = cm(t₂-t₁) ................. Equation 3

Where c = specific heat capacity of water, m = mass of water, t₁ = initial temperature of water, t₂ = final temperature of water.

substitute equation 2 into equation 3

Pt = cm(t₂-t₁) ............... Equation 4

make t the subject of the equation

t = cm(t₂-t₁)/P................ Equation 5

Given: P = 100 W, m = 1.3 kg, t₁ = 25 °C, t₂ = 100 °C.

Constant: 4200 J/kg.K

Substitute into equation 5

t = 1.3(4200)(100-25)/100

t = 409500/100

t = 4095 seconds

t = (4095/60) minutes = 68.25 minutes.

Hence the time it will take the water to reach boiling point = 68.25 minutes

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The Gulf Stream off the east coast of the United States can flow at a rapid 3.9 m/s to the north. A ship in this current has a c
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Answer:

72.54 degree west of south

Explanation:

flow = 3.9 m/s north

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to find out

point due west from the current position

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so it become like triangle with 3.3 point down and the hypotenuse is 11

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hypotenuse ×cos(angle) = adjacent side

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A rocket moves upward, starting from rest with an acceleration of 25.4 m/s^2 for 3.39 s. It runs out of fuel at the end of the 3
kolbaska11 [484]

Explanation:

Initial speed of the rocket, u = 0

Acceleration of the rocket, a=25.4\ m/s^2

Time taken, t = 3.39 s

Let v is the final velocity of the rocket when it runs out of fuels. Using the equation of kinematics as :

v=u+at

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Let x is the initial position of the rocket. Using third equation of kinematics as :

v^2=u^2+2ax_o

x_o=\dfrac{v^2}{2a}

x_o=\dfrac{86.10^2}{2\times 25.4}=145.92\ m  

Let x_o is the position at the maximum height. Again using equation of motion as :

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Now a=-g and v and u will interchange

u^2=2g(x-x_o)

x=x_o+\dfrac{u^2}{2g}

x=145.92+\dfrac{(86.10)^2}{2\times 9.8}

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