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Alexxandr [17]
3 years ago
15

How long does it take the sun to melt a block of ice at 0∘c with a flat horizontal area 1.0 m2 and thickness 1.8 cm ? assume tha

t the sun's rays make an angle of 33 ∘ with the vertical and that the emissivity of ice is 0.050.how long does it take the sun to melt a block of ice at 0∘c with a flat horizontal area 1.0 m2 and thickness 1.8 cm ? assume that the sun's rays make an angle of 33 ∘ with the vertical and that the emissivity of ice is 0.050?
Chemistry
1 answer:
VLD [36.1K]3 years ago
4 0
This is a problem involving heat transfer through radiation. The solution to this problem would be to use the formula for heat flux.

ΔQ/Δt = (1000 W/m²)∈Acosθ

A is the total surface area:
A = (1 m²) + 4(1.8 cm)(1m/100 cm)(√(1 m²))
A = 1.072 m²

ΔQ is the heat of melting ice.
ΔQ = mΔHfus
Let's find its mass knowing that the density of ice is 916.7 kg/m³.
ΔQ = (916.7 kg/m³)(1 m²)(1.8 cm)(1m/100 cm)(<span>333,550 J/kg)
</span>ΔQ = 5,503,780 J

5,503,780 J/Δt = (1000 W/m²)(0.05)(1.072 m²)(cos 33°)
<em>Δt = 122,434.691 s or 34 hours</em>

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1 mol super cooled liquid water transformed to solid ice at -10 oC under 1 atm pressure.
Arada [10]

Answer:

Explanation:

Given that:

number of moles of super cooled liquid water = 1

Melting enthalpy of ice = 6020 J/mol

Freezing point =0 °C = (0 + 273 K)= 273 K

The decrease in entropy of the system during freezing for 1 mol (i.e during transformation from liquid water to solid ice )  = - 6020 J/mol × 1 mol /273 K = -22.051 J/K

Entropy change during further cooling from 0 °C (273 K) to -10 °C (263 K)

\Delta \ S = \int\limits^{T_2}_{T_1}\dfrac{nC_p(s)dT}{T}

\Delta \ S = {nC_p(s)In \dfrac{T_2}{T_1}

\Delta \ S = {(1*37.7)In \dfrac{263}{273}

Δ S = -1.4 J/K

Total entropy change of the system = - 22.05 J/K - 1.4 J/K = - 23.45 J/K

Entropy change of universe = entropy change of the system+ entropy change of the surrounding

According to the second law of thermodynamics

Entropy change of universe  >0

SO,

Entropy change of the system + entropy change in the surrounding > 0

Entropy change in the surrounding > - entropy change of the system

Entropy change in the surrounding > - (- 23.53 J/K)

Entropy change in the surrounding > 23.53 J/K

b) Make some comments on entropy changes from the obtained data.

From the data obtained; we will realize that the entropy of the system decreases as cooling takes place when water is be convert to ice , randomness of these molecules reduces and as cooling proceeds , hence, entropy reduces more as well and the liberated heat will go into the surrounding due to this entropy of the surrounding increasing.

4 0
3 years ago
5.) A spring with a spring constant of 500 n/m is stretched 1 meter in
Ymorist [56]

Answer:

250J

Explanation:

Given parameters:

Spring constant  = 500N/m

Extension  = 1m

Unknown:

Elastic potential energy  = ?

Solution:

The elastic potential energy of a body is the energy stored within an elastic string.

  EPE  = \frac{1}{2} k e²

k is the spring constant

e is the extension

 Now;

      EPE  =  \frac{1}{2}  x 500 x 1²   = 250J

4 0
3 years ago
what happens to the density of an object when the volume of that object increases and the mass remains the same?
Dima020 [189]

Answer:

d=m/v       If v increases then the density decreases.   Denominator get  larger

Explanation:

6 0
3 years ago
We could avoid a large increase in temperature if greenhouse emissions peaked by the year _____.
Naya [18.7K]

Answer:

2036

Explanation:

  • Greenhouse gases also called GHGs are those gases found on the earth that is a product of the earth absorption and radiation of heat energy from the sun that creates a heating effect some of them are methane, carbon water vapor, ozone and nitrogen oxides.
  • The atmosphere of other planets also consists of greenhouse gases, but the impact of humans on the earth has created these gases to a certain extent that global temperatures are raised to dangerous levels.  
  • Thought the peak greenhouse gas effects and global warming process have already started in 2010, the impact of the industrial revolution has increased the atmospheric carbon content to about 50 % since the ancient times from 280 to 415 ppm in 2019.
  • At present the temperature has increased to 3 degrees from earlier times, the <u>United Nations IPCC claims that the upper limit which is to be avoided at all costs is until the year 2036</u>. Which is marked as a significant year for the upper atmospheric changes
4 0
4 years ago
The correct name for Cu(CN), is: *
garik1379 [7]

Answer:

Copper dicyanide CUPRIC CYANIDE Copper cyanide (Cu(CN)2) Cyanure de cuivre copper (II) cyanide More...

Explanation:

HOPE THIS HELPS :)

8 0
3 years ago
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