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Andrew [12]
3 years ago
10

Water ia boled at 1 at pressure in a coffe aker equpped with an immension-type electric heating element. The coffee maker intial

ly contains 1 kg of water. Once boiling started, it is observed that half of he water in the coffee maker evaporated in 10 minutes the heat loss from the cofee maker is negigible, the power rating of the heating element is: a)-1.9kw b)-16 kw c)-0.8 kW d)-2.2 kw e)-3.8KW
Physics
1 answer:
tresset_1 [31]3 years ago
4 0

-- Water cannot be boled.

-- There is no such thing as a coffe aker.

-- There is no such thing as an immension-type heating element.

Be that as it may, and it very likely still is, as it were . . .

-- The latent heat of vaporization of water at 100°C is about 2250 kilo-joules per kg.

-- To evaporate half of the kg of water in the coffee maker requires 1125 kJ of heat energy.

-- To supply that amount of heat energy over a period of 10 minutes (600 seconds), it must be supplied at a rate of

(1,125,000 Joules / 600 seconds) = 1,875 joules/second.

-- That's 1,875 watts, or <em>1.875 kilowatts</em>.

-- <em>Choice-a</em> is the choice when the solution is rounded.

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The kinetic energy of the merry-go round after 3.62  = ?

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Step 1:  Finding the Mass of merry-go-round

m = \frac{ weight}{g}

m = \frac{745}{9.81 }

m = 76.02 kg

Step 2: Finding the Moment of Inertia of solid cylinder

Moment of Inertia of solid cylinder I =0.5 \times m \times r^2

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=>0.5 \times 76.02 \times (1.45)^2

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KE = 0.5 \times I \times \omega^2

KE =0.5 \times 79.91 \times 3.69^2

KE = 544J

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