A rod made from a particular alloy is heated from 23.5°C to the boiling point of water. Its length increases by 8.21 10-4 m. The rod is then cooled from 23.5°C to the freezing point of water. By how much does the rod shrink?

Question

A rod made from a particular alloy is heated from 23.5°C to the boiling point of water. Its length increases by 8.21 10-4 m. The rod is then cooled from 23.5°C to the freezing point of water. By how much does the rod shrink?

I will be happy to critique your work.

use a proportion.

deltaL/8.21E-4=(23.5/(100-23.5)

Answer 1

To find out how much the rod shrinks when cooled from 23.5°C to the freezing point of water, you can use a proportion.

Let's represent the change in length of the rod as ΔL. The given change in temperature is from 23.5°C to the freezing point of water, so we can use the relationship:

(ΔL / 8.21E-4 m) = (23.5°C / (100°C - 23.5°C))

Now, we can solve for ΔL by cross-multiplying the equation:

ΔL = (8.21E-4 m) * (23.5°C / (100°C - 23.5°C))

Simplifying the equation further:

ΔL = (8.21E-4 m) * (23.5 / 76.5)

Calculating the value:

ΔL = 0.000821 m * 0.307

ΔL ≈ 0.000252 m

Therefore, the rod shrinks by approximately 0.000252 meters when cooled from 23.5°C to the freezing point of water.

Answer 2

To determine how much the rod shrinks when cooled from 23.5°C to the freezing point of water, you can use a proportion. The proportion should relate the change in length of the rod to the change in temperature.

Let's assume the change in length of the rod when heated from 23.5°C to the boiling point of water is given by ΔL1, and the change in length when cooled from 23.5°C to the freezing point of water is given by ΔL2.

Using the given information, we have:
ΔL1 = 8.21 x 10^-4 m (increase in length while heating)
Temperature change while heating: 100°C (boiling point) - 23.5°C (initial temperature) = 76.5°C

Now, we need to find the temperature change while cooling, which is from 23.5°C to the freezing point of water (0°C).

To find ΔL2, we can set up a proportion:

ΔL2 / ΔT1 = ΔL1 / ΔT2

Where:
ΔT1 = 76.5°C (temperature change while heating)
ΔT2 = 23.5°C - 0°C = 23.5°C (temperature change while cooling)

Now, we can plug in the values and solve for ΔL2:

ΔL2 / 76.5 = 8.21 x 10^-4 / 23.5

Cross-multiplying to solve for ΔL2:

ΔL2 = (8.21 x 10^-4) * (76.5 / 23.5)

Calculating this expression, we find that ΔL2 is approximately equal to 2.67 x 10^-4 m.

Therefore, the rod shrinks by approximately 2.67 x 10^-4 m when cooled from 23.5°C to the freezing point of water.

A rod made from a particular alloy is heated from 23.5°C to the boiling point of water. Its length increases by 8.21 10-4 m. The rod is then cooled from 23.5°C to the freezing point of water. By how much does the rod shrink?

Great

To find out how much the rod shrinks when cooled from 23.5°C to the freezing point of water, you can use a proportion.

To determine how much the rod shrinks when cooled from 23.5°C to the freezing point of water, you can use a proportion. The proportion should relate the change in length of the rod to the change in temperature.