A hollow aluminum cylinder 19.0 cm deep has an internal capacity of 2.000 L at 23.0°C. It is completely filled with turpentine at 23.0°C. The turpentine and the aluminum cylinder are then slowly warmed together to 91.0°C. (The average linear expansion coefficient for aluminum is 2.4 x10^-5/°C, and the average volume expansion coefficient for turpentine is 9.0 x10^-4/°C.)
(a) How much turpentine overflows?
in cm^3?
(b) What is the volume of turpentine remaining in the cylinder at 91.0°C? (Give you answer to four significant figures.)
in cm^3?
(c) If the combination with this amount of turpentine is then cooled back to 23.0°C, how far below the cylinder's rim does the turpentine's surface recede?
in cm?

a. 112.608 cm³ b. 2010 cm³ c. 0.032 cm

Explanation:

Given

Volume of aluminium cylinder = 2.0 L= 2.0 × 10³ cm³

Length of aluminium cylinder = 19.0 cm  

Volume of turpentine = 2.0 L= 2.0 10³ cm

Initial temperature of aluminium cylinder = 23.0 °C

Initial temperature of turpentine = 23.0 °C

Average linear expansion coefficient of aluminium α = 2.4 × 10⁻⁵ °C

Average volume expansion coefficient of turpentine γ₁ = 9.0 × 10⁻⁴ °C

(a) How much turpentine overflows in cm³?

We calculate the volume expansion of the aluminium cylinder and that of the turpentine and then subtract the difference.

The volume expansion of material  V = V₀(1 + γΔθ)

where V = final volume, V₀ = initial volume, γ = volume expansion and Δθ = temperature change.

For aluminium V₀ = 2000 cm³, γ = 3α = 3 × 2.4 × 10⁻⁵ /°C = 7.2 × 10⁻⁵ /°C

Δθ = θ₂ - θ₁ = 91 °C - 23 °C = 68 °C

V₁ = 2000(1 + 7.2 × 10⁻⁵ /°C × 68°C) = 2000(1 + 0.004896) = 2000(1.004896) = 2009.792 cm³

For turpentine V₀ = 2000 cm³, γ = 9.0 × 10⁻⁴ /°C

Δθ = θ₂ - θ₁ = 91 °C - 23 °C = 68 °C

V₂= 2000(1 + 9.0 × 10⁻⁴ /°C × 68°C) = 2000(1 + 0.0612) = 2000(1.0612) = 2122.4 cm³.

The volume of turpentine that overflows = V₂ - V₁ = 2122.4 cm³ - 2009.792 cm³ = 112.608 cm³

(b) What is the volume of turpentine remaining in the cylinder at 91.0°C? (Give you answer to four significant figures.)  in cm³?

The volume of turpentine remaining is the final volume of the turpentine minus overflow = 2122.4 cm³ - 112.608 cm³ = 2009.792 cm³ = 2010 cm³ to four significant figures.

(c) If the combination with this amount of turpentine is then cooled back to 23.0°C, how far below the cylinder's rim does the turpentine's surface recede in cm?

The linear expansion for the aluminium to 91 C from 23 C is L = L₀(1 + Δθ)

For aluminium L₀ = 19 cm, α = 2.4 × 10⁻⁵ /°C =

Δθ = θ₂ - θ₁ = 91 °C - 23 °C = 68 °C

L = 19(1 + 2.4 × 10⁻⁵ /°C × 68°C) = 19(1 + 0.001632) = 19(1.001632) = 19.031 cm

The linear expansion for the aluminium from 91 C to 23 C is L = L₀(1 + Δθ)

For aluminium L₀ = 19 cm, α = 2.4 × 10⁻⁵ /°C =

Δθ = θ₂ - θ₁ = 23 °C - 91 °C = -68 °C

L₁ = 19.031(1 + 2.4 × 10⁻⁵ /°C × -68°C) = 19.031(1 - 0.001632) = 19.031(0.998368) = 18.999 cm

How far below the rim it recedes is L - L₁ = 19.031 cm - 18.999 cm = 0.032 cm.

i think they are all correct

explanation:

they seem good

e/c, where e is its total energy