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kirill [66]
3 years ago
9

How much heat is needed to boil 120 kg of water ?

Physics
1 answer:
nekit [7.7K]3 years ago
7 0
Q = mc<span>∆t, where:
q = energy flow
m = mass, 120 000 g
c = specific heat capacity, 4.81 J/gC
</span><span>∆t = change in temperature, ~75 (100 - 25, which is room temperature)

Substituting in the values, we get:
q = 120000 x 4.81 x 75 = 43290000 Joules = 43.29 MJ

Hope I helped!! xx


</span>
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juin [17]
  • A 16.0 kg canoe moving to the left at 12.5 m/s makes an elastic head on collision with a 14.0 kg raft moving to the right at 16.0 m/s.
  • After the collision the raft moves to the left at 14.4 m/s assuming water simulates a frictionless surface.
  • Mass of the canoe (m1) = 16 Kg
  • Mass of the raft (m2) = 14 Kg
  • Initial velocity of the canoe (u1) = 12.5 m/s
  • Initial velocity of the raft (u1) = - 16 m/s [Here, the raft's velocity is negative, because the objects are moving in the opposite direction]
  • Total momentum of the system = m1u1 + m2u2 = [(16 × 12.5) + (14 × -16)] Kg m/s = (200 - 224) Kg m/s = -24 Kg m/s
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  • Let the final velocity of the canoe be v1.
  • Total momentum of the system after the impact = m1v1 + m2v2 = [(16 × v1) + (14 × 14.4)] Kg m/s = 16v1 Kg + 201.6 Kg m/s
  • According to the law of conservation of momentum, Total momentum of the system before the impact = Total momentum of the system after the impact
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<u>Answer:</u>

<u>T</u><u>he final velocity of the </u><u>canoe </u><u>is </u><u>-</u><u>1</u><u>4</u><u>.</u><u>1</u><u> </u><u>m/</u><u>s </u><u>or </u><u>1</u><u>4</u><u>.</u><u>1</u><u> </u><u>m/</u><u>s </u><u>to </u><u>the </u><u>right.</u>

Hope you could get an idea from here.

Doubt clarification - use comment section.

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Answer:

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We have given speed = 72.4 miles per hour

We have to convert this speed in meter per second

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We also know that 1 km = 1000 m

So 115.84 km =115.84\times 10^3m

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