Question

Suppose a mercury thermometer contains 3.250g of mercury and has a capillary that is 0.180mm in diameter.. How far does the mercury rise in the capillary when the temperature changes from 0.0 ∘C to 25.0 ∘C? The density of mercury at these temperatures is 13.596 g/cm3 and 13.534 g/cm3, respectively.

Answer 1

The mercury will rise in the capillary to a height of 4.3 cm.

FURTHER EXPLANATION

The steps in solving this problem are the following:

1. Sort the given from the problem.

2. Get the volume of mercury at 0.0 °C and 25.0 °C.

3. Get the difference in volume for the two temperatures.

4. Solve for the height of the mercury using the volume of a cylinder (shape of the thermometer).

STEP 1. Sort the given information in the problem.

Given:

mass of mercury = 3.250 g

diameter of capillary = 0.180 mm

density at 0.0°C = 13.596 g/cm³

density at 25.0° = 13.534 g/cm³

Find:

volume change from 0.0°C to 25.0° C

change in height from 0.0°C to 25.0°C

STEP 2: Using the equation for density, calculate the volume at 0.0°C and 25.0° C. The general equation to be used is:

$volume = \frac{mass}{density}$

For 0.0°C

$volume = \frac{3.250 \ g}{13.596 \frac{g}{cm^3}}\\\boxed {volume = 0.23904 \ cm^3}$

For 25.0°C

$volume = \frac{3.250 \ g}{13.534 \frac{g}{cm^3}}\\\boxed {volume = 0.24014 \ cm^3}$

STEP 3: Solve for the volume change.

$\Delta V = 0.24014 \ cm^3 - 0.23904 \ cm^3\\\boxed {\Delta V = 1.1 \times 10^{-3} \ cm^3}$

STEP 4: From the change in volume, the change in height can be calculated using the equation below:

$\Delta h = \frac{\Delta V}{\pi\ r^2}\\\Delta h = \frac{1.1 \times 10^{-3} \ cm^3}{\pi({9 \times 10^{-3} \ cm)}^2}\\\\\boxed {\Delta h = 4.32 \ cm}$

From the given in the problem, the least number of significant figures is 2, thus, the final answer must only have 2 significant figures as well.

Therefore,

$\boxed {\boxed {\Delta h = 4.3 \ cm}}$

LEARN MORE

1. Learn more about thermal expansion brainly.com/question/1166774
2. Learn more about capillary action brainly.com/question/1295312

Keywords: mercury thermometer

Answer 2

1) You need to get volume of both temperatures by using first attached formula V= Mass/Density$V_1 = (3.250 g)/(13.596 g/(cm^3)) = .2390 cm^3 At 25 degrees C (V_2e have: V_2 = (3.250 g)/(13.534 g/(cm^3)) = .2401 cm^3$

2) Using the second formula you get the height of 0 degree 
$h = V/(pi * r^2) h_1 =(.2390 cm^3)/(pi*(.009 cm)^2) = 939.2 cm$

(radius in cm is $(0.180 mm) / 2 * (1 cm)/(10 mm) or .009 cm$

3) Then with h1 you can easily get the height of 25 degrees 
$h_2 =(.2401 cm^3)/(pi*(.009 cm)^2) = 943.5 cm$

 Subtract 943.5 cm - 939.2 cm, and obtain a rise in mercury height of 4.3 cm