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dusya [7]
3 years ago
7

Water does not run out of dropper unless the rubber-bulb is pressed. Why?

Physics
1 answer:
nekit [7.7K]3 years ago
8 0

Answer:

The pressure inside the dropper is same as the atmospheric pressure when the rubber bulb is not pressed. ... But when we press the rubber bulb the pressure inside the dropper increases and hence the water flows out.  Atmospheric pressure acting from outside the dropper balances the pressure exerted by water and water does not come out of a dropper. On pressing the dropper inside pressure of water becomes more than outside atmospheric pressure and water run out. When we press the bulb of a dropper with its nozzle kept in water, air in the dropper is seen to escape in the form of bubbles. Once we release the pressure on the bulb, water gets filled in the dropper.

                                             <u><em>Thank You</em></u>

                                        <u> </u>Please mark me brainliest

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Calculate the amount of heat needed to raise 1.0 kg of ice at -20 degrees Celsius to steam at 120 degree Celsius
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Answer:

801.1 kJ

Explanation:

The ice increases in temperature from -20 °C to 0 °C and then melts at 0 °C.

The heat required to raise the ice to 0 °C is Q₁ = mc₁Δθ₁ where m =  mass of ice = 1 kg, c₁ = specific heat capacity of ice = 2108 J/kg°C and Δθ₁ = temperature change. Q₁ = 1 kg × 2108 J/kg°C × (0 - (-20))°C = 2108 J/kg°C × 20  °C = 4216 J

The latent heat required to melt the ice is Q₂ = mL₁ where L₁ = specific latent heat of fusion of ice = 336000 J/kg. Q₁ = 1 kg × 336000 J/kg = 336000 J

The heat required to raise the water to 100 °C is Q₃ = mc₂Δθ₂ where m =  mass of ice = 1 kg, c₂ = specific heat capacity of water = 4187 J/kg°C and Δθ₂ = temperature change. Q₃ = 1 kg × 4187 J/kg°C × (100 - 0)°C = 4187 J/kg°C × 100  °C = 418700 J

The latent heat required to convert the water to steam is Q₄ = mL₂ where L = specific latent heat of vapourisation of water = 2260 J/kg. Q₄ = 1 kg × 2260 J/kg = 2260 J

The heat required to raise the steam to 120 °C is Q₅ = mc₃Δθ₃ where m =  mass of ice = 1 kg, c₃ = specific heat capacity of steam = 1996 J/kg°C and Δθ₃ = temperature change. Q₃ = 1 kg × 1996 J/kg°C × (120 - 100)°C = 1996 J/kg°C × 20  °C = 39920 J

The total amount of heat Q = Q₁ + Q₂ + Q₃ + Q₄ + Q₅ = 4216 J + 336000 J

+ 418700 J + 2260 J + 39920 J = 801096 J ≅ 801.1 kJ

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