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lesya [120]
4 years ago
12

The largest building in the world by volume is the boeing 747 plant in Everett, Washington. It measures approximately 632 m long

, 710 yards wide, and 112 ft high.
What is the cubic volume in feet, convert your result from part a to cubic meters?
Physics
1 answer:
Schach [20]4 years ago
3 0

Explanation:

Given that,

The dimensions of the largest building in the world is 632 m long, 710 yards wide, and 112 ft high. It basically forms a cuboid. The volume of a cuboidal shape is given by :

Since,

1 meter = 3.28084 feet

632 m = 2073.49 feet

1 yard= 3 feet

710 yards = 2130 feet

V = lbh

V=2073.49 \ ft\times 2130\ ft\times 112\ ft

V=494651774.4\ ft^3

V=4.94\times 10^8\ ft^3

Also,

V=(4.94\times 10^8\ ft^3)(\dfrac{1\ m}{3.281})^3

V=1.39\times 10^7\ m^3

Hence, this is the required solution.

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To solve this problem we will apply the concept related to the heat transferred to a body to reach a certain temperature. This concept is shaped by the energy ratio of a body which is the product of the mass, its specific heat and the change in temperature. For the specific case, it will be the sum of the heat transferred to the Water, the Aluminum and the loss due to latency due to vaporization in the water. That is to say,

\Delta Q = m_{Al} C_p \Delta T +m_wC_w \Delta T  +m_w L_v

Here,

m_{Al}= Mass of Aluminum

C_p= Specific Heat of Aluminum

C_w= Specific Heat of Water

m_w = Mass of water

L_v =Latent of Vaporization

Replacing,

\Delta Q = (0.85)(900)(100-45)+(2)(2000)(100-45)+(0.7)(2258000)

Converting,

\Delta Q = 1842675J (\frac{0.000239006kCal}{1J})

\Delta Q = 440.409kCal

Therefore is required 440.409kCal

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Read 2 more answers
A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.15 m2 and whose thickn
KengaRu [80]

Answer:

88 %

Explanation:

The rate of heat loss by a conducting material of thermal conductivity K, cross-sectional area,A and thickness d with a temperature gradient ΔT is given by

P = KAΔT/d

The total heat lost by the styrofoam wall is P₁ = K₁A₁ΔT₁/d₁ where K₁ =thermal conductivity of styrofoam wall 0.033 W/m-K, A₁ = area of styrofoam wall = 17 m², ΔT₁ = temperature gradient between inside and outside of the wall and d₁ = thickness of styrofoam wall = 0.20 m

The total heat lost by the glass window is P₂ = K₂A₂ΔT₂/d₂ where K₂ =thermal conductivity of glass window pane wall 0.96 W/m-K, A₂ = area of glass window pane = 0.15 m², ΔT₂ = temperature gradient between inside and outside of the window and d₂ = thickness of glass window pane = 7 mm = 0.007 m

The total heat lost is P = P₁ + P₂ = K₁A₁ΔT₁/d₁ + K₂A₂ΔT₂/d₂

Now, since the temperatures of both inside and outside of both window and wall are the same, ΔT₁ = ΔT₂ = ΔT

So, P = K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂

Since P₂ = K₂A₂ΔT₂/d₂ = K₂A₂ΔT/d₂is the heat lost by the window, the fraction of the heat lost by the window from the total heat lost is

P₂/P = K₂A₂ΔT/d₂ ÷ (K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂)

= 1/(K₁A₁ΔT/d₁÷K₂A₂ΔT/d₂ + 1)

= 1/(K₁A₁d₂÷K₂A₂d₁ + 1)

= 1/[(0.033 W/m-K × 17 m² × 0.007 m ÷ 0.96 W/m-K × 0.15 m² × 0.20 m) + 1]

= 1/(0.003927/0.0288 + 1)

= 1/(0.1364 + 1)

= 1/1.1364

= 0.88.

The percentage is thus P₂/P × 100 % = 0.88 × 100 % = 88 %

The percentage of heat lost by window of the total heat is 88 %

6 0
3 years ago
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