THERMAL PROPERTIES OF MATTER

1. A pie plate is filled up to the brim with pumpkin pie filling. The pie plate is made of Pyrex and its expansion can be neglected. It is a cylinder with an inside depth of 2.10 cm and an inside diameter of 30.0cm. It is prepared at a room temperature of 68 degrees Fahrenheit and placed in an oven at 400 degrees Fahrenheit. When it’s taken out, 151cc of the pie filling has flowed out and over the rim. Determine the coefficient of volume expansion of the pie filling, assuming it is a fluid.

Answer:

The volume expanded is 151 cc. the volume of reference, that is, the V of cylinder is 0.25*3.1415*2.10*(30.0) ^2the coefficient of volume expansion= 151/ VThe expansion of the Pyrex plate (container) is neglected.So V (final) =V (Initial)=V of the cylinder.

2. At the bottom of an old mercury-in-glass thermometer is a 46.0-mm3 resevoir filled with mercury. When the thermometer was placed under your tongue, the warmed mercury would expand into a very narrow cylindrical channel, called a capillary, whose radius was 2.50 x 10-2 mm. Marks were placed along the capillary that indicated the temperature. Ignore the thermal expansion of the glass and determine how far (in mm) the mercury would expand into the capillary when the temperature changed by 1.0 C°

Answer:

The change in the mercury's volume is equal to beta*V*delta T, where beta is the coefficient given above, V is the original volume, and delta T is the temperature change. Since the diameter of the capillary is unchanged, the change in volume there will be pi*r-squared*delta L, where r is the radius, and deltaL is the change in the length of the mercury column. Setting the volume changes equal to each other, and doing a little algebra,beta*V*deltaT = pi*r-squared*deltaLdeltaL = beta*V*deltaT / pi*r-squareddeltaL = 1.8x10(-4)*46*1.0 / 3.14*[2.5x10(-2)]*[2.5x10(-2)]deltaL = 4.21 mm

3. The coefficient of volume expansion for ethyl alcohol is 1.12x10^-3 1/K. Copper has a coefficient of volume expansion of 1.67x10^-3 1/K. If a copper cylinder is filled to the brim with ethyl alcohol at 5 C, what percentage of the alcohol will spill if the temperature is raised by 25 C.

Answer:

The temperature change is 20 C. So, the volume expansion of the copper will be20x1.67x10^-3 or 3.34x10^-2 liters and the volume expansion of the alcohol will be 20x1.12x10^-3=2.24x10^-2 l

4. The density of gasoline is 730 kg/m3 at 0°C. Its average coefficient of volume expansion is 9.60 x 10-4/C°. If 1.00 gallon of gasoline occupies 0.00380 m3, how many extra kilograms of gasoline would you get if you bought 8.0 gal of gasoline at 0°C rather than at 19.0°C from a pump that is not temperature compensated?

Answer:

Density of gasoline at 0°C = 730 kg/m3Density of gasoline at 19°C = 730 (1 + 19(9.60 x 10-4)) kg/m3 = 743.3152 kg/mm=dVV = 8 gal = 0.0304 m3difference in m = 743.3152(0.0304) - 730(0.0304) kg= 0.40478208 kg

5. A pair of eyeglass frames are made of an epoxy plastic (coefficient of linear expansion = 1.30 10-4°C-1). At room temperature (20.0°C), the frames have circular lens holes 2.10 cm in radius. To what temperature must the frames be heated if lenses 2.11 cm in radius are to be inserted into them?

ANSWER:

ΔL = αLoΔT(.0211 - .0210) = (1.3e-4)(.021)(T - 20).0001 = (2.73e-6)(T - 20)36.63003663 = T - 20T = 56.63003663T = 56.6°C

The frames must be heated to 56.6°C for the frames to increase from a 2.1cm radius to a 2.11cm radius.

6. The aptly-named Steel Bridge over the Willamette River in Portland, Oregon, has a central span steel truss that is 64.1 m long. What is its change in length over the year from the average minimum annual temperature of 0.90°C to the average maximum of 26.3°C? Use a coefficient of linear expansion of 1.17×10^−5 (1/C°).

Answer:

delta L = length chg of central truss = (L)(Cx) (delta T) = (64.1)(1.17e-5)(26.3 - 0.9)delta L = 75e-5(25.4) = 1905e-5 = 0.0191 m = 1.91 cm

7. We cut a circular piece from the rectangular plate. Which ones of the processes given below can help us in passing through the circular piece from the hole?

Answer:

I. Increasing the temperatures of rectangular plate and circular piece
II. Decreasing the temperature of the circular piece
III. Decreasing the temperatures of the rectangular plate and circular piece

I. If we increase the temperatures of the plate and circular piece, expansion of the hole and the circular piece will be the same. Thus, this option can help us.
II. If we decrease the temperature of the circular piece, it contracts and hole becomes larger than the piece. This option can also help us.
III. If we decrease the temperatures of the plate and circular piece, hole and circular piece contract in same size. This process can also help us.

8. There are three same metal rods having same length and thickness. If the temperatures of them are given like; T, 2T and 3T find the relations of final lengths of the rods.

Answer:

We find the final temperatures of the system by the formula;
Tfinal=T1+T2+T3/3=6T/3=2T
Since the temperature of the first rod increase, its final length also increases. Temperature of the second rod stays same, thus there won’t be change in the length of this rod. Finally, temperature of the third rod decreases, thus its contract and final length of it decreases with respect to initial length. As a result relation of the final lengths of the rods;
L1>L2>L3