Calculating the coefficent of volume expansion of liquid?

AI Thread Summary
The discussion focuses on calculating the coefficient of volume expansion for a liquid in a thermometer, which increased in volume by 40% with a temperature change of 15 degrees Celsius. The formula ΔV = VβΔT is suggested for finding the coefficient, with a calculation of 0.4/15 yielding approximately 0.0267. Participants discuss how to determine the increase in volume and the radius of the cylindrical stem based on the known volume increase and the liquid's rise. One user confirms that the explanation provided was clear and helpful, leading to a successful resolution of the problem. Ultimately, the original poster expresses gratitude for the assistance received.
Pinchie81
Messages
1
Reaction score
0
I can't seem to figure out this question.

A Thermometer is filled with unknown liquid. It has a square bulb and a stem with cylinder shape. It is found that the volume of the liquid increased by 40% when the temperature changes by 15 Celsius.
Calculate the coefficient of expansion of volume of the liquid and the radius of the stem if the liquid rises by 5 cm?
Do I need to use the formula: ΔV = VβΔT

Don't know how to tackle this question any help would be appreciated.
 
Last edited:
Physics news on Phys.org
Pinchie81 said:
I can't seem to figure out this question.

A Thermometer is filled with unknown liquid. It has a square bulb and a stem with cylinder shape. It is found that the volume of the liquid increased by 40% when the temperature changes by 15 Celsius.
Calculate the coefficient of expansion of volume of the liquid and the radius of the stem if the liquid rises by 5 cm?
Do I need to use the formula: ΔV = VβΔT

Don't know how to tackle this question any help would be appreciated.

WOW! I am stuck on exactly same question.
Someone tried to help me with this..I thought I understand it all now but I am stuck again haha.
This might make sense to you then you can help me.
For 1st part the volume coefficient of expansionis the fractional increase in volume per unit rise in temperature. The % increase is given so change that to a fraction and divide by the rise in temperature (0.4/15 = 0.0267?? may be :S)
For the 2nd part if you know the vol of the bulb, then you can work out the increase in the actual vol of the liquids because you know it increases by 40%. The tube is a cylinder and liquid will have a vol = area of cross section x length. You know the increase in volume and you know the length so work out the area of cross section and hencethe radius.
 
SAFiiNA said:
This might make sense to you then you can help me.
For 1st part the volume coefficient of expansionis the fractional increase in volume per unit rise in temperature. The % increase is given so change that to a fraction and divide by the rise in temperature (0.4/15 = 0.0267?? may be :S)
For the 2nd part if you know the vol of the bulb, then you can work out the increase in the actual vol of the liquids because you know it increases by 40%. The tube is a cylinder and liquid will have a vol = area of cross section x length. You know the increase in volume and you know the length so work out the area of cross section and hencethe radius.
I think you explained that very well, SAFiiNA. And judging by the absence of any further query from Pinchie81, I'd conclude that you managed to answer his/her questions! :wink:
 
NascentOxygen said:
I think you explained that very well, SAFiiNA. And judging by the absence of any further query from Pinchie81, I'd conclude that you managed to answer his/her questions! :wink:

Thank you! :) I finally managed to work it out myself.
 
Thread ''splain this hydrostatic paradox in tiny words'
This is (ostensibly) not a trick shot or video*. The scale was balanced before any blue water was added. 550mL of blue water was added to the left side. only 60mL of water needed to be added to the right side to re-balance the scale. Apparently, the scale will balance when the height of the two columns is equal. The left side of the scale only feels the weight of the column above the lower "tail" of the funnel (i.e. 60mL). So where does the weight of the remaining (550-60=) 490mL go...
Consider an extremely long and perfectly calibrated scale. A car with a mass of 1000 kg is placed on it, and the scale registers this weight accurately. Now, suppose the car begins to move, reaching very high speeds. Neglecting air resistance and rolling friction, if the car attains, for example, a velocity of 500 km/h, will the scale still indicate a weight corresponding to 1000 kg, or will the measured value decrease as a result of the motion? In a second scenario, imagine a person with a...
Scalar and vector potentials in Coulomb gauge Assume Coulomb gauge so that $$\nabla \cdot \mathbf{A}=0.\tag{1}$$ The scalar potential ##\phi## is described by Poisson's equation $$\nabla^2 \phi = -\frac{\rho}{\varepsilon_0}\tag{2}$$ which has the instantaneous general solution given by $$\phi(\mathbf{r},t)=\frac{1}{4\pi\varepsilon_0}\int \frac{\rho(\mathbf{r}',t)}{|\mathbf{r}-\mathbf{r}'|}d^3r'.\tag{3}$$ In Coulomb gauge the vector potential ##\mathbf{A}## is given by...
Back
Top