Solving Density ρ Question: Δρ = -87x10-6ρ at 25ºC to -40ºC

  • Thread starter Thread starter A_Munk3y
  • Start date Start date
  • Tags Tags
    Density
AI Thread Summary
The discussion focuses on calculating the change in density (Δρ) of a lead sphere as its temperature decreases from 25°C to -40°C, using the formula Δρ = -βρΔT. The value of β for lead is given as 87x10^-6. The user initially misinterprets the question by calculating Δρ without considering the fractional change in density, which is required. The correct approach involves dividing the change in density by the original density to find the fractional change. The conversation highlights the importance of understanding the specific requirements of the problem.
A_Munk3y
Messages
72
Reaction score
0

Homework Statement


Show that the change in the density ρ of a substance, when the temperature changes by ΔT,
is given by Δρ = −βρ ΔT. (b) What is the fractional change in density of a lead sphere whose
temperature decreases from 25°C to −40º C?

Homework Equations


βLead= 87x10-6

The Attempt at a Solution


ok i already go the first part, but i don't think I am doing b right.

im just plugging in (-40-25) for temperature, and the β of lead.
so Δρ = −87x10-6ρ
ρ=176.8Δρ

i doubt that's the answer. am i misunderstanding the question here? or am i just doing something completely wrong?
 
Physics news on Phys.org
A_Munk3y said:
ρ=176.8Δρ

This is right, but the question asks for fractional change in density. That means the change in density divided by the density itself.
 
oh... ok
thank you :)
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top