Thermodynamics expansion problem

AI Thread Summary
The problem involves calculating the depth of a swimming pool that overflows when water warms from 24 °C to 34 °C. The volume expansion coefficient for water is given as 2.07 x 10^-4 °C−1. The initial attempt incorrectly equated the change in depth (1.2 cm) with the change in volume. Clarification revealed that the change in volume should be calculated using the volume expansion formula, and the correct approach does not require taking the cube root. The user successfully found the correct answer after receiving guidance.
xnitexlitex
Messages
22
Reaction score
0

Homework Statement


On a hot summer day, a cubical swimming pool is filled to within 1.2 cm of the top with water at 24 °C. When the water warms to 34 °C, the pool overflows. What is the depth of the pool? (The volume expansion coefficient for water is 2.07 10-4 °C−1.)


Homework Equations


ΔV = βV0ΔT


The Attempt at a Solution


I set ΔV equal to 1.2, plugged the numbers in, got 193.24 cm3 for the volume, and took the cube root of the result for the depth. That didn't work. What am I supposed to do?
 
Physics news on Phys.org
Hi xnitexlitex! :smile:

ΔV is not equal to 1.2.
1.2 cm is the change in depth, not the change in volume.

How did you plug your numbers in and get your volume?
Because I see no reason why you would need to take a cube root...
 
Thank you. I just got the answer now.
 

Thread 'Struggling to understand the concept of this question (ice cube in water optics question)'

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...
View full post »
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
View full post »

Similar threads

Why Is Volume Excluded in My Thermal Expansion Calculation?
Replies
1
Views
2K
Iron Ring Expansion: Solve for Δθ to Find Temp Increase
Replies
2
Views
2K
Volume expansion and sea level rise
Replies
6
Views
2K
Solving Volumetric Expansion of Water in Glass
Replies
6
Views
2K
Amount of water spilled when the temperature is changed
Replies
9
Views
2K
Adiabatic Expansion of Pressurized Air in a Piston-Cylinder Setup
Replies
8
Views
2K
Linear and Volume expansion Problem
Replies
2
Views
8K
Problem: using thermal expansion to calculate sea level rise
Replies
13
Views
7K
Coeff. of Linear Exp. of glass from vol expansion of content
Replies
6
Views
7K
What is the correct solution for the temperature expansion problem?
Replies
3
Views
1K

Hot Threads

Recent Insights

Back
Top