Exercicios Sobre Dilatação Linear

Linear Expansion Exercises: What Actually Works

Most people mess up linear thermal expansion because they treat it like a plug-and-chug topic. It isn't. You need to understand what the equation is actually telling you before you start crunching numbers, and even then the setup matters more than the math. The core equation is straightforward: L = L · · T. Change in length equals original length times the coefficient of linear expansion times the temperature change. That's it. But the trick is in knowing which L to use, which applies, and whether the problem is asking for final length or just the change.

exercicios sobre dilatação linear para fixar o conteúdo

Here's a realistic set of problems I've assigned over the years, plus walkthroughs of the ones where students consistently lose points.

Problem 1: Basic Application

A steel rail is 12 meters long at 20°C. What is its length at 45°C? The coefficient of linear expansion for steel is approximately 12 × 10 °C¹. Start by identifying your knowns: L = 12 m, T_initial = 20°C, T_final = 45°C, so T = 25°C. Now plug in: L = 12 × 12 × 10 × 25. Multiply step by step. 12 times 12 is 144. 144 times 25 is 3600. So L = 3600 × 10 = 0.0036 meters, or 3.6 millimeters. The final length is 12.0036 m. The rail doesn't grow much, but on a 100-meter bridge section, that adds up fast.

Problem 2: The Trap

This one trips people up. A brass rod measures exactly 50.00 cm at 0°C. An aluminum rod measures 50.10 cm at 0°C. At what temperature will both rods have the same length? _brass = 19 × 10 °C¹ and _aluminum = 23 × 10 °C¹. Set up the equation: L_brass = L_aluminum at the unknown temperature T. So 50.00(1 + 19 × 10 × T) = 50.10(1 + 23 × 10 × T). Expand both sides: 50.00 + 50.00 × 19 × 10 × T = 50.10 + 50.10 × 23 × 10 × T. Do the multiplication: 50.00 + 0.00095T = 50.10 + 0.0011523T. Rearrange: 0.00095T - 0.0011523T = 50.10 - 50.00. That gives -0.0002023T = 0.10. Solve: T -494°C.

So you need to cool them down almost 500 degrees. The aluminum contracts faster, so at sufficiently low temperatures the shorter brass rod actually ends up longer when heated, but at this low temperature the aluminum shrinks enough to catch up. Makes physical sense if you check it.

Problem 3: Compound Bars

A bimetallic strip is made of steel and copper bonded together. _steel = 12 × 10 °C¹, _copper = 17 × 10 °C¹. Both strips are 30 cm at 20°C. The strip is heated to 120°C. By how much does the copper exceed the steel in length change? Calculate each separately: L_steel = 0.30 × 12 × 10 × 100 = 0.00036 m = 0.36 mm. L_copper = 0.30 × 17 × 10 × 100 = 0.00051 m = 0.51 mm. The difference is 0.15 mm. That small gap is exactly what makes the strip bend, which is why bimetallic strips work as thermostats.

Common Mistakes I See in Grading

Students forget that L is always the original length at the reference temperature, not some random value pulled from the text. If a problem says a structure is X meters long at 30°C and then asks for the change when heated to 50°C, your L is still X — you don't subtract the 30 from anything. The formula already handles the temperature difference separately. Another thing: unit consistency. If is given in °C¹, your T should be in °C. If it's in K¹, your T should be in K. Since the magnitude of one degree Celsius equals one kelvin, the numerical value of T is the same either way, but mixing up the units in the coefficient itself (some tables give values in different temperature scales) will derail your answer.

I once had a student lose points on an exam because he used the volumetric expansion coefficient instead of the linear one. The problem said = 36 × 10 °C¹ for a material and asked about a wire. He treated it as linear and got the wrong order of magnitude. Turns out the 36 was (volumetric), and the linear coefficient would have been 12 × 10. Always double-check which coefficient a problem gives you. Some textbooks are sloppy about notation.

Practice Problems

Try these on your own. Answers follow. 1. An iron bridge is 200 m long at 10°C. Find the expansion at 40°C. _iron = 12 × 10 °C¹.

2. A glass bottle is filled to the brim with 750 mL of water at 20°C. What volume overflows when heated to 85°C? _water = 210 × 10 °C¹, _glass = 27 × 10 °C¹. 3. A steel tape measure calibrated at 20°C is used to measure a distance at 35°C. The tape reads 50.00 m. What is the actual distance? _steel = 11 × 10 °C¹.

4. A copper ring has an inner diameter of 4.000 cm at 20°C. A steel ball has a diameter of 4.010 cm at 20°C. To what temperature must the ring be heated so the ball just passes through? _copper = 17 × 10 °C¹, _steel = 12 × 10 °C¹. 5. Two rods, one of aluminum (L = 80 cm) and one of steel (L = 100 cm), are both at 0°C. Their coefficients are 23 × 10 °C¹ and 12 × 10 °C¹ respectively. Find the temperature at which the difference in their lengths becomes 0.10 cm.

Answers

1. L = 200 × 12 × 10 × 30 = 0.072 m = 7.2 cm. 2. V_water = 750 × 210 × 10 × 65 = 10.2375 mL. V_glass = 750 × 27 × 10 × 65 = 1.31625 mL. Overflow = 10.2375 - 1.31625 8.92 mL.

3. The tape expands, so each meter marked on the tape is actually longer than a real meter. The scale is stretched. Actual distance = 50.00 × (1 + 11 × 10 × 15) = 50.00 × 1.000165 = 50.00825 m. The reading underestimates the true distance by about 8.3 mm. 4. Set 4.000(1 + 17 × 10 × T) = 4.010(1 + 12 × 10 × T). Solving gives T 294°C.

5. The length difference is (100 - 80) + (100 × 12 × 10 × T - 80 × 23 × 10 × T) = 20 + T(0.0012 - 0.00184) = 20 - 0.00064T. Set equal to 19.90 (since the difference should decrease by 0.10 cm): T = 20/0.00064 31,250°C. That's clearly unrealistic, which means the aluminum never catches up to the steel within any physically meaningful temperature range. The initial 20 cm gap is too large relative to the small difference in expansion coefficients. This is a good reminder to sanity-check your answers. The hardest part of these problems isn't the algebra. It's setting up the right equation and knowing what each variable represents. Once you get comfortable with that, the calculations themselves are trivial arithmetic. I'd recommend doing at least ten of these before moving on to area and volume expansion — the linear version builds the intuition you need for everything else.