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adelina 88 [10]
3 years ago
6

What is its diameter when the temperature is raised to 100 degrees Celsius? (b) What temperature change is required to increase

its volume
Physics
1 answer:
alexandr1967 [171]3 years ago
4 0

The question is incomplete, the complete question is;

A spherical steel ball bearing has a diameter of 2.540cm at 25.00∘C.

(a) What is its diameter when it's temperature is raised to

100∘C?

(b) What temperature change is required to increase its volume by

1.000% ?

Answer:

a) 2.542 cm

b) 303.03°C

Explanation:

Given;

Diameter of the ball= 2.540cm

Initial temperature= 25.0°C

Final temperature= 100.0°C

Percentage increase in volume = 1.000%

Temperature coefficient of expansion for steel =11.0×10^−6/∘C

d2= d1[1 + α(T2-T1)]

d2= 2.540[1 + 11.0×10^−6(100-25)]

d2= 2.540[1 + 8.25×10^-4]

d2= 2.542 cm

From;

%V ×1/100 = V ×3α ×∆T/ V

Substituting values;

1.000 ×1/100= 3× 11.0×10^−6 × ∆T

∆T= 0.01/3× 11.0×10^−6

∆T= 303.03°C

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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

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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

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<u>T₂ = 49.3°C</u>

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