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sergey [27]
3 years ago
9

A styrofoam container used as a picnic cooler contains a block of ice at 0°C. If 325 g of ice melts in 1 hour, how much heat ene

rgy per second is passing through the walls of the container? The heat of fusion of ice is 3.33 x 10^5 J/kg. Answer in units of W.
Physics
1 answer:
Sauron [17]3 years ago
8 0

Answer:

30.0625 W

Explanation:

325 g/h   x    (1h x 1kg)/(3600s x 1000g)   x   3,33 x 10^5 J/Kg = 30.0625 J/Kg = 30.0625 W

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A narrow copper wire of length L and radius b is attached to a wide copper wire of length L and radius 2b, forming one long wire
Morgarella [4.7K]

Answer:

electric field in the wide wire is

E₂ =\frac{E}{4}

Explanation:

given

length of the copper wire = L

radius of the copper wire r₁ = b

length of the second copper wire = L

radius of the second copper wire r₂ = 2b

electric field in the narrow wire = E₁=E

recall

resistance R = ρL/A

where ρ is resistivity of the copper wire, L is the length, and A is the cross sectional area.

Resistance of narrow wire, R₁

R₁ = ρL/A

where A  = πb²

R₁ = ρL/πb²---------- eqn 1

Resistance of wide wire, R₂

R₂ = ρL/A

where A = π(2b)²

R₂ = ρL/π(2b)²

R₂ = ρL/4πb²-------------- eqn 2

R₂ = ¹/₄(ρL/πb²)

comparing eqn 1 and 2

R₁ = 4R₂

calculating the current in the wire,

I = E/(R₁ + R₂)

recall

R₁ = 4R₂

∴ I = E/(4R₂ + R₂)

I = E/5R₂

calculating the potential difference across R₁ & R₂

V₁ = IR₁

I = E/5R₂

∴ V₁ = ER₁/5R₂

R₁ = 4R₂

V₁ = 4ER₂/5R₂

∴V₁  = ⁴/₅E

potential difference for R₂

V₂= IR₂

I = E/5R₂

∴ V₂ = ER₂/5R₂

V₂ = ER₂/5R₂

∴V₂  = ¹/₅E

so, electric field E = V/L

for narrow wire E₁ = V₁/L ----------- eqn 3

for wide wire, E₂ = V₂/L------------ eqn 4

compare eqn 3 and 4

E₂/E₁ = V₂/V₁( L is constant)

E₂/E₁ = ¹/₅E/⁴/₅E

E₂ = E₁/4

note E₁ = E

∴E₂ =\frac{E}{4}

8 0
3 years ago
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