How Do Temperature Changes Affect Physical Properties in Basic Physics Problems?

[FestiveF]

Basic lower level physics questions...

Hi all! I am new to the forum and I have a few lower level physics problem with which I am confused. Any help and explanations will be immensely appreciated!

1) A steel pipeline is 1500 m long and is at a temperature of 30 degrees Celcius. How much shorter will it be if the temperature falls to 10 degrees Celcius?
*I got .36 m...

2) How much heat is absorbed by 255 g of methanol when it is heated from 10 degrees Celcius to 60 degrees Celcius?
*I got 3.12 x 10^4 J...

3) Calculate the number of kJ of heat energy liberated by 4 kg of steam at 100 degrees when it is condensed, cooled, and changed to ice at 0 degrees.
* I got 1.2 x 10^4 kJ...

4) If the density of mercury is 13.59 x 10^3 kg/m^3 at 20 degrees Celcius, what will its density be at 65 degrees Celcius?

5) In a certain hydraulic lift, the small piston has a radius of 15 cm. The large piston has a radius of 26 cm. A force of 225 N is applied to the small piston. What is the mass of the crate that is being lifted by the application of this 225 N force?

Thank you SO much ahead of time ~

 

[quantumdude]

Hi FestiveF,

I am closing this thread because it is identical to a thread you posted in the Physics forum. Interested readers can join the discussion here:

https://www.physicsforums.com/showthread.php?s=&threadid=3453

 

[M98Ranger]

1) To find the change in length, we can use the formula: ΔL = αLΔT, where α is the coefficient of linear expansion and L is the original length. We are given the temperature change (ΔT = 30 - 10 = 20 degrees Celcius) and the original length (L = 1500 m), but we need to find the coefficient of linear expansion for steel. According to Google, the coefficient of linear expansion for steel is 11.7 x 10^-6 m/m-°C. Plugging in the values, we get: ΔL = (11.7 x 10^-6 m/m-°C)(1500 m)(20°C) = 0.351 m. Therefore, the steel pipeline will be 0.351 m shorter when the temperature falls from 30°C to 10°C.

2) The formula for heat energy is Q = mcΔT, where m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. We are given the mass (m = 255 g), the change in temperature (ΔT = 60 - 10 = 50°C), and the specific heat capacity of methanol (c = 2.51 J/g-°C). Plugging in the values, we get: Q = (255 g)(2.51 J/g-°C)(50°C) = 3.18 x 10^4 J. Therefore, 255 g of methanol will absorb 3.18 x 10^4 J of heat energy when heated from 10°C to 60°C.

3) The formula for heat energy is Q = ml, where m is the mass, and l is the latent heat of fusion/vaporization. We are given the mass (m = 4 kg) and the latent heat of vaporization for steam (l = 2.26 x 10^6 J/kg). To find the heat energy liberated, we need to convert the latent heat from J/kg to kJ/kg (since the mass is given in kg and we want the final answer in kJ). Therefore, we get: Q = (4 kg)(2.26 x 10^6 J/kg) = 9.04 x 10^6 J = 9.04 x 10^3 kJ. Therefore, 4