An aluminum can is filled to the brim with a liquid. The can and the liquid are heated so their temperatures change by the same amount. The can’s initial volume at 8 °C is 4.8 × 10-4 m3. The coefficient of volume expansion for aluminum is 69 × 10-6 (C°)-1. When the can and the liquid are heated to 80 °C, 6.1 × 10-6 m3 of liquid spills over. What is the coefficient of volume expansion of the liquid?
The volume of liquid that spills over the can is the difference between the increase in the volume of the liquid and that of the aluminum can
ΔV=β₁VΔT - β ₂VΔT
β₁=(ΔV/VΔT) + β ₂ =
=(6.1•10⁻⁶/4.8•10⁻⁴•72) + 69•10⁻⁶ = 1.13•10⁻³ °C⁻¹
Did the liquid spill because the aluminum can couldn't handle the heat and collapsed like a house of cards? That would be quite a sight to see! Anyway, let me help you with the question.
To solve this problem, we need to find the change in volume of the aluminum can. It is given that the volume of the can changes by 6.1 × 10-6 m3 when the temperature changes by the same amount as the liquid, which is 80 °C - 8 °C = 72 °C.
The change in volume of the can can be calculated using the coefficient of volume expansion for aluminum:
ΔV_can = V_initial * β_aluminum * ΔT,
where ΔV_can is the change in volume of the can, V_initial is the initial volume of the can, β_aluminum is the coefficient of volume expansion for aluminum, and ΔT is the change in temperature.
Plugging in the values:
6.1 × 10-6 m3 = 4.8 × 10-4 m3 * (69 × 10-6 (C°)-1) * 72 °C,
Simplifying:
6.1 × 10-6 m3 = 4.8 × 10-4 m3 * 69 * 10-6 * 72 °C,
Dividing both sides by (4.8 × 10-4 m3 * 69 × 10-6 * 72 °C):
1 = β_aluminum * ΔT,
Finally, rearranging the equation to solve for the coefficient of volume expansion of the liquid, β_liquid:
β_liquid = 1 / ΔT,
β_liquid = 1 / 72 °C.
So, the coefficient of volume expansion of the liquid is approximately 0.01 (C°)-1.
To solve this problem, we need to use the principle of thermal expansion and the concept of conservation of volume.
Let's assume the initial volume of the liquid inside the can is V.
1. Calculate the change in temperature.
The change in temperature is given by the final temperature minus the initial temperature:
ΔT = 80 °C - 8 °C = 72 °C
2. Calculate the change in volume of the can.
The change in volume of the can can be calculated using the coefficient of volume expansion for aluminum. It is given by:
ΔV_can = V_can * β_al * ΔT
where V_can is the initial volume of the can, β_al is the coefficient of volume expansion for aluminum, and ΔT is the change in temperature.
Substituting the values, we get:
ΔV_can = (4.8 × 10^(-4) m^3) * (69 × 10^(-6) (°C)^(-1)) * (72 °C)
3. Calculate the final volume of the can.
The final volume of the can can be calculated by adding the change in volume of the can to its initial volume:
V_can_final = V_can_initial + ΔV_can
Substituting the values, we get:
V_can_final = (4.8 × 10^(-4) m^3) + [(4.8 × 10^(-4) m^3) * (69 × 10^(-6) (°C)^(-1)) * (72 °C)]
4. Calculate the volume of the liquid that spills over.
The volume of the liquid that spills over is given as 6.1 × 10^(-6) m^3.
5. Calculate the final volume of the liquid inside the can.
The final volume of the liquid inside the can can be calculated by subtracting the volume of the liquid that spills over from the final volume of the can:
V_liquid_final = V_can_final - V_liquid_spilled
where V_liquid_spilled is the volume of the liquid that spills over.
Substituting the values, we get:
V_liquid_final = [(4.8 × 10^(-4) m^3) + [(4.8 × 10^(-4) m^3) * (69 × 10^(-6) (°C)^(-1)) * (72 °C)]] - (6.1 × 10^(-6) m^3)
6. Calculate the coefficient of volume expansion of the liquid.
The coefficient of volume expansion of the liquid can be calculated using the following equation:
β_liquid = ΔV_liquid / (V_liquid * ΔT)
where ΔV_liquid is the change in volume of the liquid, V_liquid is the initial volume of the liquid, and ΔT is the change in temperature.
Substituting the values, we get:
β_liquid = [(V_liquid_final - V_liquid) / (V_liquid * ΔT)]
/ (V_liquid / (V_liquid * ΔT))
Simplifying, we get:
β_liquid = (V_liquid_final - V_liquid) / V_liquid
Now we can substitute the known values and calculate the coefficient of volume expansion of the liquid.
To find the coefficient of volume expansion of the liquid, we need to use a relationship involving the change in volume, initial volume, coefficient of volume expansion, and change in temperature.
First, let's find the change in volume of the aluminum can when it is heated from 8 °C to 80 °C.
ΔV_can = V_final_can - V_initial_can
where ΔV_can is the change in volume of the can, V_final_can is the final volume of the can at 80 °C, and V_initial_can is the initial volume of the can at 8 °C.
We can calculate the change in volume of the can as follows:
ΔV_can = V_final_can - V_initial_can
ΔV_can = 4.8 × 10^(-4) m^3 x 69 x 10^(-6) (°C)^(-1) x (80 - 8) °C
ΔV_can = 4.8 × 10^(-4) m^3 x 69 x 10^(-6) (°C)^(-1) x 72 °C
ΔV_can = 2.84 x 10^(-9) m^3
Next, we know that 6.1 × 10^(-6) m^3 of liquid spills over when the can and liquid are heated to 80 °C. The change in volume of the liquid is equal to the volume that spilled over:
ΔV_liquid = 6.1 × 10^(-6) m^3
Since both the can and liquid experienced the same change in temperature, we can equate their change in volume:
ΔV_can = ΔV_liquid
Therefore, 2.84 × 10^(-9) m^3 = 6.1 × 10^(-6) m^3
Now, we can solve for the coefficient of volume expansion of the liquid (β_liquid):
β_liquid = ΔV_liquid / (V_initial_can x ΔT)
where β_liquid is the coefficient of volume expansion of the liquid, V_initial_can is the initial volume of the can, and ΔT is the change in temperature.
Plugging in the values:
β_liquid = (6.1 × 10^(-6) m^3) / ((4.8 × 10^(-4) m^3) x (80 - 8) °C)
β_liquid = (6.1 × 10^(-6) m^3) / ((4.8 × 10^(-4) m^3) x 72 °C)
β_liquid ≈ 1.04 × 10^(-3) (°C)^(-1)
Therefore, the coefficient of volume expansion of the liquid is approximately 1.04 × 10^(-3) (°C)^(-1).