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Ipatiy [6.2K]
3 years ago
15

Calculate the heat flux (in W/m^2) through a sheet of a metal 14 mm thick if the temperatures at the two faces are 350 and 140°C

. Assume steady-state heat flow and that the thermal conductivity of this metal is 52.4 W/m-K. (b) What is the heat loss per hour (in J/h) if the area of the sheet is 0.42 m2? (c) What will be the heat loss per hour (in J/h) if a material with a thermal conductivity of 1.8 W/m-K is used? (d) Calculate the heat loss per hour (in J/h) if the first metal is used and the thickness is increased to 24 mm.
Physics
1 answer:
ExtremeBDS [4]3 years ago
5 0

Answer:

Explanation:

The rate of conductive heat transfer in watts is:

q = (k/s) A ΔT

where k is the heat conductivity, s is the thickness, A is the area, and ΔT is the temperature difference.

a)

Given k = 52.4 W/m/K, s = 0.014 m, and ΔT = 350-140 = 210 K, we can find q/A:

q/A = (52.4 / 0.014) (210)

q/A = 786,000 W/m²

b)

Given that A = 0.42 m², we can find q:

q = (0.42 m²) (786,000 W/m²)

q = 330,120 W

A watt is a Joule per second.  Convert to Joules per hour:

q = 330,120 J/s * 3600 s/hr

q = 1.19×10⁹ J/hr

c)

If we change k to 1.8 W/m/K:

q = (k/s) A ΔT

q = (1.8 / 0.014) (0.42) (210)

q = 11,340 J/s

q = 4.08×10⁷ J/hr

d)

If k is 52.4 W/m/K and s is 0.024 m:

q = (k/s) A ΔT

q = (52.4 / 0.024) (0.42) (210)

q = 192,570 J/s

q = 6.93×10⁸ J/hr

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A 7950-kg railroad car travels alone on a level frictionless track with a constant speed of 15.0 m/s . A 2950-kg load, initially
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The new speed of car is 10.9 m/s

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According to the principle of momentum conservation, momentum is only modified by the action of forces as they are outlined by Newton's equations of motion; momentum is never created nor destroyed inside a problem domain.

Mass of the railroad car, m₁ = 7950 kg

Mass of the load, m₂ = 2950 kg

It can be assumed as the speed of the car, u₁ = 15 m/s

Initially, it is at rest, u₂ = 0

Let v is the speed of the car. It can be calculated using the conservation of momentum as :

m_1u_1 + m_2u_2 = (m_1 + m_2) v

v =\frac{m_1u_1}{m_1+m_2}

v = \frac{7950*15}{7950+2950}

v= 10.9 m/s

Therefore, the new speed of care is 10.9 m/s

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An 75-kg hiker climbs to the summit of Mount Mitchell in western North Carolina. During one 2.0-h period, the climber's vertical
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Answer:

The change in gravitational potential energy of the climber-Earth system is  \Delta  PE  = 396900 \ J

Explanation:

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    The time  taken is  T  =  2 \ hr =  2 *  3600 =  7200 \ s

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        \Delta  PE  =  75  *  9.8  *  540

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How many protons, electrons, and neutrons are present in an atom of cr-24?
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The number of protons, electrons, and neutrons in the atom of cr-24 is 28, 24, and 24 respectively.

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An atom is the smallest unit of matter, consisting of the positively charged nucleus and the electrons which move around it. The atom can not be divided further.

The atom of a matter is made by three elements-

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The atom of cr-24 has the atomic number 24. The atomic number of atom is equal to the number of proton and electron in an atom. Thus,

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Now, if this atom has the mass number of 52 (most common). Then the number of neutron is,

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Thus, the number of protons, electrons, and neutrons in the atom of cr-24 is 28, 24, and 24 respectively.

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