A diver dives 30m deep into a shallow sea floor. What will the change in pressur

Question

A diver dives 30m deep into a shallow sea floor. What will the change in pressure be when the diver reaches sea floor? 1 kg mass of ice is heated from -50 C to 110C. What is the energy required for this process? Solve the diagram using Bermoulli's equation What happens to the velocity, if the radius is shrunk by 1/2? What is the relation between velocity, pressure and radius of a tube for the flow of viscous fluid using Poiseuille's equation, where the coefficient of Viscosity and A = pi R^2, then the velocity will be:

Explanation / Answer

we are allowed to answer only 1 question at a time.

1)
rho, density of sea water = 1029 Kg/m^3

change in pressure = rho*g*h
= 1029 * 9.8*30
= 302526 Pa
Answer: 302526 Pa

2)
ice to steam:
specific heat capacity of water, C1 = 4.186 J/goC
specific heat capacity of ice, C2 = 2.09 J/goC
specific heat capacity of steam, C3 = 2.01 J/goC
Latent heat of fusion of ice, Lf = 334 J/g
latent heat of vaporization of water LV = 2264.76 J/g
m=1kg = 1000 g

heat required to take ice from -50 oC to 0 oC,
Q1 = m*C2*delta T
=1000*2.09*(50)
= 9718.5 J

heat required to convert ice to water,
Q2 = m* Lf
= 1000*334
= 334000 J

heat required to take water from 0 oC to 100 oC,
Q3 = m*C1*delta T
=1000*4.186*(100-0)
= 418600 J

heat required to convert water to steam,
Q4 = m* LV
= 1000*2264.76
= 2264760 J

heat required to take steam from 100 oC to 110 oC,
Q5 = m*C3*delta T
=1000*2.01*(110-100)
= 20100 J

Total heat required= Q1 + Q2 + Q3 + Q4 + Q5
= 9718.5 + 334000 + 418600 + 2264760 + 20100
= 3047179 J
= 304.7 KJ
Answer: 304.7 KJ

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