Coeff. of Linear Exp. of glass from vol expansion of content

[Qwurty2.0]

Homework Statement

The coefficient of volume expansion of olive oil is 0.68 × 10-3 K-1. A 1-liter glass beaker is
filled to the brim with olive oil at room temperature. The beaker is placed on a range and the
temperature of the oil and beaker increases by 25 C°. As a result, 0.0167 liters of olive oil spill
over the top of the beaker. What is the coefficient of linear expansion of glass?

Homework Equations

ΔV = βV_iΔT
ΔV = (3α)V_iΔT
β = Coefficient of Volume Expansion
α = Coefficient of Linear Expansion

The Attempt at a Solution

V_i = 1.0 L (for both olive oil and the beaker)
T_i = 20°C = 293.15 K
ΔT = 25°C = 298.15 K
V_lost = 0.0167 L
β_olive = 0.68x10^-3 K^-1
α_glass = ?

Calculate volume expansion of beaker
ΔV_olive = (0.68x10^-3 K^-1) * (1.0 L) * (298.15 K)
= 0.202742 L

ΔV_glass = ΔV_olive - V_lost
= 0.202742 L - 0.0167 L
= 0.186042 L (Volume increase for glass beaker)

Calculate Coefficient of Linear Expansion for Glass Beaker
0.186042 L = (3α_glass)(1.0 L)(298.15 K)
0.186042 L = 894.45 L⋅K * α_glass
α_glass = 0.186042 L / (894.45 L⋅K)
= 0.000208 / K = 2.1x10^-4 K^-1

This is very close to one of the actually answers (there are 5). One of them is 2x10^-5 K^-1. The teacher has said that he has removed the correct answer for some of the questions, and that we are to explain why are answer is correct if that is the case. Is my answer correct? If not, what did I do wrong?

 

[SteamKing]

Hint: think significant figures. Does your result carry the correct number of significant figures?

 

[Qwurty2.0]

I believe so. There is 1 significant figure (the 1 liter volume), so the answer I would have is 2x10^-4.

But the issue isn't the sig. figs. (at least I don't think), it's the exponent. My answer has 2x10^-4, while the closest answer is 2x10^-5. I've gone over my calculations and I believe I have done it correctly.

 

[SteamKing]

I believe so. There is 1 significant figure (the 1 liter volume), so the answer I would have is 2x10^-4.

But the issue isn't the sig. figs. (at least I don't think), it's the exponent. My answer has 2x10^-4, while the closest answer is 2x10^-5. I've gone over my calculations and I believe I have done it correctly.

You're getting confused on how to deal with temperatures in your calculations.

Remember, degrees Kelvin and degrees Celsius use the same temperature scale, but each scale has a different starting point. For the change in temperature ΔT of 25 °C, you have assumed that this is the same as a temperature of 298.15 °K, which is wildly incorrect for calculating a change in volume.

 

[Qwurty2.0]

But if 20°C (293.15 K) is the initial temperature and the final temperature is T_i + ΔT = 20°C + 25°C = 293.15 K + 298.15 K = 591.3 K, and ΔT is T_f - T_i, then isn't ΔT = 591.3 K - 293.15 K = 298.15 K? What am I missing? Don't they give the change in temp right away? :(

 

[SteamKing]

But if 20°C (293.15 K) is the initial temperature and the final temperature is T_i + ΔT = 20°C + 25°C = 293.15 K + 298.15 K = 591.3 K, and ΔT is T_f - T_i, then isn't ΔT = 591.3 K - 293.15 K = 298.15 K? What am I missing? Don't they give the change in temp right away? :(

You're treating a temperature reading and a change in temperature as the same thing, which is not correct.

If an initial temperature reading is 20 °C and the temperature increases by 25 °C, then the final temperature is 45 °C, which is not equivalent to 591.3 °K. You can check this by converting °K to °C. Remember, a temperature reading of 0 °C on the Celsius scale is the same as a temperature reading of 273.15 °K on the Kelvin scale, but a change in temperature of 25 °C is the same as a change in temperature of 25 °K.

On both the Celsius and Kelvin temperature scales, the difference in temperature between the freezing point of water and the boiling point of water is 100°.

 

[Qwurty2.0]

Ah, my mistake. I understand what I was doing wrong. Thank you.