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3 years ago

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m_Cu * sh_CuA system consists of a copper tank whose mass is 13 kilogram , 4 kilogram of liquid water, and an electrical resisto

r of negligible mass. The system is insulated on its outer surface. Initially, the temperature of the copper is 27 degC and the temperature of the water is 50 degC . The electrical resistor transfers 100 kilojoule to the system. Eventually the system comes to equilibrium. Determine the final equilibrium temperature, in ∘C.

1 answer:

notsponge [240]3 years ago

8 0

Answer:

T₂ = 49.3°C

Explanation:

Applying law of conservation of energy to the system we get the following equation:

Energy Supplied by Resistor = Energy Absorbed by Tank + Energy Absorbed by Water

E = mC(T₂ - T₁) + m'C'(T'₂ - T'₁)

where,

E = Energy Supplied by Resistor = 100 KJ = 100000 J

m = mass of copper tank = 13 kg

C = Specific Heat of Copper = 385 J/kg.°C

T₂ = Final Temperature of Copper Tank

T₁ = Initial Temperature of Copper Tank = 27°C

T'₂ = Final Temperature of Water

T'₁ = Initial Temperature of Water = 50°C

m' = Mass of Water = 4 kg

C' = Specific Heat of Water = 4179.6 K/kg.°C

Since, the system will come to equilibrium finally. Therefor: T'₂ = T₂

Therefore,

(100000 J) = (13 kg)(385 J/kg.°C)(T₂ - 27°C) + (4 kg)(4179.6 J/kg.°C)(T₂ - 50°C)

100000 J = (5005 J/°C)T₂ - 135135 J + (16718.4 J/°C)T₂ - 835920 J

100000 J + 135135 J + 835920 J = (21723.4 J/°C)T₂

(1071055 J)/(21723.4 J/°C) = T₂

<u>T₂ = 49.3°C</u>

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Answer:

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Explanation:

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Where:

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faust18 [17]

Hello!

Use the <u>second law of Newton:</u>

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Replacing:

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Resolving:

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Air in a 120 km/h wind strikes head on the face of abuilding 45m wide by 75m high and is brought to rest. if air has a mass of 1

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From conservation of momentum, the ram force can be calculated similarly to rocket thrust:
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Mary and her younger brother Alex decide to ride the carousel at the State Fair. Mary sits on one of the horses in the outer sec

Alex

Answer:

ωM = 1.7 rad/s

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ωA = 1.7 rad/s

vA = 1.7 m/s

Explanation:

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We use the following formula for circular motion:

ω = θ / t Formula (1)

Where:

ω: angular speed (rad/s)

θ:angle that the particle travels (rad)

t: time interval (s)

Equivalence

1 revolution = 2π rad

Data

θ= 1 revolution = 2π rad

t = 3.7 s

We replace data in the formula (1):

ω = (2π rad) / (3.7s )= 1.7 rad/s

Because the angular speed depends only on the number of turns in a time interval then Mary's angular speed (ωM) is equal to Alex's angular speed (ωA).

ωM = ωA = 1.7 rad/s

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We use the following formula for circular motion:

v = ω*r Formula (2)

Where:

v : tangential speed (m/s)

ω: angular speed ( rad/s)

r : radius of the circular path (m)

Data

ωM = ωM = ω= 1.7 rad/s

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rA = 1m : Alex's circular path radius

We apply the formula (2) to calculate v:

vM = ω* rM = (1.7) (2) = 3.4 m/s

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