At an absolute temperature T0, a cube has sides of length L0...

At an absolute temperature $T_{0},$ a cube has sides of length $L_{0}$ and has density $\rho_{0}$ . The cube is made of a material with coefficient of volume expansion $\beta$ (a) Show that if the temperature increases to $T_{0}+\Delta T$ , the density of the cube becomes approximately $$\rho \approx \rho_{0}(1-\beta \Delta T)$$ (Hint I Lise the expression $(1+x)^{n} \approx 1+n x$ , valid for $|x| \ll 1 . )$ Explain why this approximate result is valid only if $\Delta T$ is much less than $1 / \beta,$ and explain why you would expect this to be the case in most situations. (b) A copper cube has sides of length 1.25 $\mathrm{cm}$ at $20.0^{\circ} \mathrm{C}$ . Find the change in its volume and density when its temperature is increased to $70.0^{\circ} \mathrm{C}$ .

At an absolute temperature $T_{0},$ a cube has sides of length $L_{0}$ and has density $\rho_{0}$ . The cube is made of a material with coefficient of volume expansion $\beta$ (a) Show that if the temperature increases to $T_{0}+\Delta T$ , the density of the cube becomes approximately
$$\rho \approx \rho_{0}(1-\beta \Delta T)$$
(Hint I Lise the expression $(1+x)^{n} \approx 1+n x$ , valid for $|x| \ll 1 . )$ Explain why this approximate result is valid only if $\Delta T$ is much less than $1 / \beta,$ and explain why you would expect this to be the case in most situations. (b) A copper cube has sides of length 1.25 $\mathrm{cm}$ at $20.0^{\circ} \mathrm{C}$ . Find the change in its volume and density when its temperature is increased to $70.0^{\circ} \mathrm{C}$ .

Key Concepts

Binomial Approximation

The binomial approximation, given by (1+x)^n ? 1+nx for |x| much smaller than 1, is a mathematical tool used to simplify expressions involving powers of sums. In the context of thermal expansion, it allows for an approximate treatment of expressions relating volumetric changes to temperature changes when such changes are relatively small, thereby simplifying the calculations for density and volume variations.

Mass Conservation in Thermal Processes

When a solid object such as a cube is heated, its volume may change due to thermal expansion, but its mass remains constant. Since density is defined as mass per unit volume, a change in volume while maintaining constant mass will result in an inverse change in density. Understanding this relationship is crucial when analyzing systems where temperature-driven volume changes affect material properties.

Validity of Linear Approximations

Linear approximations, such as using the binomial approximation, are valid when the change in the variable (in this case, temperature) is small compared to a relevant scale (here, 1/?). This concept underscores the importance of recognizing the limitations of approximations in physics; using a linear model is justified only for small perturbations where higher-order terms in the expansion are negligible.

Thermal Expansion

Thermal expansion is the tendency of matter to change in shape, area, and volume in response to a change in temperature. When a material is heated, its particles move apart, causing it to expand. This concept is fundamental in understanding how physical dimensions change with temperature, and it plays a critical role in engineering and material science applications.

Coefficient of Volume Expansion

The coefficient of volume expansion (?) quantifies how much the volume of a material changes per degree change in temperature. It is a material-specific parameter that indicates the fractional change in volume per unit temperature change when the temperature variation is small. This concept is essential for predicting and calculating dimensional changes in solids, liquids, and gases due to temperature variations.

Transcript

00:01 We know that the density of an object, row is equal to its mass over its volume.

00:07 We also know that under a temperature change, its volume changes by delta v is equal to v -0, beta -dalta -tta -t.

00:14 The delta -t is the changing temperature, and beta is the volumetric coefficient of expansion.

00:21 So if an object undergrows a temperature change from t -0 to t -0 plus delta -tt, its volume changes from v -0, to v0 plus delta v.

00:38 So under such temperature change, the density of the object becomes m over v0 plus delta v.

00:51 But we can substitute the expression for delta v from above.

00:55 So under temperature change, the density becomes m over v0 plus v0 beta delta t.

01:06 So from here we do some mathematics and divide the top and the bottom by b0.

01:13 And we get this is m over v0 all over 1 plus beta delta t.

01:21 But m over v0 is simply the original density of the object roanot before the object's temperature was changed.

01:31 That's the ro knot over 1 plus beta delta t.

01:37 So we can actually write this as the density of an object row is equal to its initial density row not into one plus beta delta t to the power minus one but then we can use a following expression one plus x also the power n is approximately one plus nx so long as x is much less than one so for very small.

02:12 X we can make that approximation and hence we can rewrite so in this case our n is minus 1 and then we can rewrite row as row not into 1 minus beta delta so what we are saying is that beta delta t is very small so this is the case if delta if delta t is much less than 1 over beta and we know 1 over beta is on the order of 10 to the 4 celsius degrees that the t in the other hand is typically on the order of about 10 to the 2 celsius degrees so this approximation is then accurate and hence we have an expression for our new density...

Please give Ace some feedback