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natita [175]
3 years ago
12

A freezer compartment is covered with a 2-mm-thick layer of frost at the time it malfunctions. If the compartment is in ambient

air at 20 °C and a coefficient of h 2 W/(m2 K) characterizes heat transfer by natural convection from the exposed surface of the layer, estimate the time required to completely melt the frost. The frost may be assumed to have a mass density of 700 kg/m3 and a latent heat of 334 kJ/kg.
Chemistry
1 answer:
ruslelena [56]3 years ago
7 0

Answer:

The time required to melt the frost is 3.25 hours.

Explanation:

The time required to melt the frost dependes on the latent heat of the frost and the amount of heat it is transfered by convection to the air .

The heat transferred per unit area can be expressed as:

q=h_c*A*\Delta T\\\\q/A=h_c*\Delta T

being hc the convective heat transfer coefficient (2 Wm^-2K^-1) and ΔT the difference of temperature (20-0=20 °C or K).

q/A=h_c*\Delta T=2\frac{W}{m^2K}*20K=40\frac{W}{m^2}

If we take 1 m^2 of ice, with 2 mm of thickness, we have this volume

V=T*A = 0.002 m * 1 m^2=0.002m^3

The mass of the frost can be estimated as

M=\rho * V=700\frac{kg}{m^3}*0.002m^3= 1.4 kg

Then,  the amount of heat needed to melt this surface (1 m²) of frost is

Q=L*M=334\frac{kJ}{kg}*1.4kg= 467.6kJ

The time needed to melt the frost can be calculated as

t=\frac{Q}{(q/A)}=\frac{467.6kJ/m2}{40W/m2} = 11.69\frac{kJ}{W}*\frac{1W*s}{1J}*\frac{1000J}{1kJ}=   11690s=3.25h

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rewona [7]

Answer:

nuclear energy

Explanation:

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3 years ago
Calculate the number of kilojoules to warm 125 g of iron from 23.5 °C to 78.0 °C.
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3024.75 Joules needed to warm iron
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3 years ago
Can someone please help me with this ASAP, this is due today, please please help me with this.
hichkok12 [17]

Answer:

this is fairly simple if you have a periodic table with you.

Explanation:

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5 0
2 years ago
1. Pewien tlenek azotu o masie cząsteczkowej 108 u zawiera 74,07% tlenu. Wykonaj stosowne obliczenia i napisz wzór sumaryczny te
love history [14]

Answer:

1. Stąd empiryczny wzór substancji to N₂O₅

2. W związku z tym ilość w gramach chlorku sodu NaCl is 114,4 g.

Explanation:

1. Mamy tutaj;

Masa molowa tlenku azotu = 108u

Masa azotu = 14,0067u

Masa tlenu = 15,999 u

74,07% masy tlenku azotu to tlen

Dlatego masa obecnego tlenu = 108 × 74,07 / 100 = 79,9956 u

Masa obecnego azotu = 108 - 79,9956 = 28,0044u

Liczba moli tlenu = 79,9956 / 15,999 = 5,00003 ≈ 5

Liczba moli azotu = 28,0044 / 14,0067 = 1,99935 ≈ 2

Stąd empiryczny wzór substancji to N₂O₅.

2. Kiedy sód reaguje z chlorem, mamy;

2Na (s) + Cl₂ (g) → 2NaCl (s)

Dlatego 2 mole sodu Na reaguje z 1 molem chloru gazowego Cl₂, z wytworzeniem 2 moli chlorku sodu NaCl

W związku z tym 1 mol sodu Na reaguje z 1/2 molem chloru gazowego Cl₂ z wytworzeniem 1 mola chlorku sodu NaCl

Masa Na obecnego w reakcji = 45 g

Masa molowa sodu = 22,989769u

Liczbę moli sodu w 45 g sodu podano w następujący sposób;

Liczba \, \, moli \, \, Na= \frac{Mass \, of \, Na}{Molowy \, masa \, z \, Na} = \frac{45}{22.989769} = 1.96 \, mole

Z czego 1,96 moli sodu Na reaguje z 1/2 × 1,96 mola chloru gazowego Cl₂ z wytworzeniem 1,96 mola chlorku sodu NaCl

Masa molowa NaCl = 58,44 g / mol

Dlatego masa NaCl = liczba moli NaCl × masa molowa NaCl

Masa NaCl = 1,96 × 58,44 = 114,39001 g ≈ 114,4 g

W związku z tym ilość w gramach chlorku sodu NaCl = 114,4 g.

4 0
3 years ago
(01.01 LC)
Vadim26 [7]
The answer to this question is B
4 0
3 years ago
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