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Help me solve this physics question number 24

Here is the solution to the question. Check the picture below 👇👇👇

Pls I am at home with no hope to eat today, I am working in a school but we have not been paid our salary...., Pls I need your help with any amount to feed myself today, even yesterday I went without food.

0632908849 (gt bank)....thank you.

Pls note; this is not because I solved the question, but I am just asking if you could help me.

NDSMELODY:
Here is the solution to the question. Check the picture below 👇👇👇

Pls I am at home with no hope to eat today, I am working in a school but we have not been paid our salary...., Pls I need your help with any amount to feed myself today, even yesterday I went without food.

0632908849 (gt bank)....thank you.

Pls note; this is not because I solved the question, but I am just asking if you could help me.

Volume "expansionity"(whatever the hell that means) is not 3 multiplied by linear "expansionity" so your solution is incorrect ab-initio. Linear coefficient of expansion is just that, the the fraction of the linear dimension of an object that must be added to the original linear dimension of aforementioned object to obtain the new linear dimension after the object is heated. To get the new volume, use the information to calculate the increase in the cubes side length, getting (L_0 + DL) where L_0 is the original side length and DL is the increase in length due to heating, then compute the volume with (L_0 + DL) ^3, cube the new length to get the new volume.

Your second mistake is that you failed to note that the heat coefficients were given in Kelvin while the problem asked for increase in volume after the cubic object is increased from 10 Celsius to 60 Celsius. Before you write even the first line of equation you must convert these temperatures to Kelvin. You can do this by adding 273K to the given temperatures, because 0C=273K

Alusiizizi:
Volume "expansionity"(whatever the hell that means) is not 3 multiplied by linear "expansionity" so your solution is incorrect ab-initio. Linear coefficient of expansion is just that, the the fraction of the linear dimension of an object that must be added to the original linear dimension of aforementioned object to obtain the new linear dimension after the object is heated. To get the new volume, use the information to calculate the increase in the cubes side length, getting (L_0 + DL) where L_0 is the original side length and DL is the increase in length due to heating, then compute the volume with (L_0 + DL) ^3, cube the new length to get the new volume.

Your second mistake is that you failed to note that the heat coefficients were given in Kelvin while the problem asked for increase in volume after the cubic object is increased from 10 Celsius to 60 Celsius. Before you write even the first line of equation you must convert these temperatures to Kelvin. You can do this by adding 273K to the given temperatures, because 0C=273K

I would like you to provide your solution and show your workings clearly.

Alusiizizi:
Volume "expansionity"(whatever the hell that means) is not 3 multiplied by linear "expansionity" so your solution is incorrect ab-initio. Linear coefficient of expansion is just that, the the fraction of the linear dimension of an object that must be added to the original linear dimension of aforementioned object to obtain the new linear dimension after the object is heated. To get the new volume, use the information to calculate the increase in the cubes side length, getting (L_0 + DL) where L_0 is the original side length and DL is the increase in length due to heating, then compute the volume with (L_0 + DL) ^3, cube the new length to get the new volume.

Your second mistake is that you failed to note that the heat coefficients were given in Kelvin while the problem asked for increase in volume after the cubic object is increased from 10 Celsius to 60 Celsius. Before you write even the first line of equation you must convert these temperatures to Kelvin. You can do this by adding 273K to the given temperatures, because 0C=273K

from the first line of your comment, I believe you have not studied expansivity very well before you made your comment if otherwise you shouldn't have commented at all...............I just have one question for you,.......is there any relationship between linear expansivity and volume expansivity, is there any relationship between linear expansivity and area expansivity??
Therefore, show mathematically the relationship between the quantities stated above ?

Alusiizizi:
Volume "expansionity"(whatever the hell that means) is not 3 multiplied by linear "expansionity" so your solution is incorrect ab-initio. Linear coefficient of expansion is just that, the the fraction of the linear dimension of an object that must be added to the original linear dimension of aforementioned object to obtain the new linear dimension after the object is heated. To get the new volume, use the information to calculate the increase in the cubes side length, getting (L_0 + DL) where L_0 is the original side length and DL is the increase in length due to heating, then compute the volume with (L_0 + DL) ^3, cube the new length to get the new volume.

Your second mistake is that you failed to note that the heat coefficients were given in Kelvin while the problem asked for increase in volume after the cubic object is increased from 10 Celsius to 60 Celsius. Before you write even the first line of equation you must convert these temperatures to Kelvin. You can do this by adding 273K to the given temperatures, because 0C=273K

This is why many students fail exams, when you do not understand questions, how will you provide correct solution to such question. The question was not based on linear expansivity, therefore how would you use linear expansion formula......just smiling, you need to do your research very well next time before commenting and in case you are a physics teacher/tutor, kindly consult your text before you teach your students non- sense !!

Mr NDSMELODY...You are the one that don't know what you are doing. Changing it to kelvin still gives you the same answer. Learn from your oga on top

Charmopompie:
Mr NDSMELODY...You are the one that don't know what you are doing. Changing it to kelvin still gives you the same answer. Learn from your oga on top

1. How does converting from one unit to another now change the answer, if you truly understand the basic as you claimed , then you don't need to argue on this .

2. I know you don't still understand the question, for backing up wrong comment up there about the relationship between linear and volume expansivity. The question was not based on linear expansivity so you can't use linear expansion formula.

3. Go and study your physics text very well before you comment.

Charmopompie:
Mr NDSMELODY...You are the one that don't know what you are doing. Changing it to kelvin still gives you the same answer. Learn from your oga on top

my brother this is education/academics...., Your oga at the top gave wrong explanation about the question

*Problem Solution*

Given:

- Side length (a) = 10 cm = 0.1 m
- Initial temperature (T1) = 10°C = 283 K
- Final temperature (T2) = 60°C = 333 K
- Linear expansivity (α) = 1.2 × 10^(-5) K^(-1)
- Volume expansivity (β) = 3α = 3.6 × 10^(-5) K^(-1) (for a cube)

Change in temperature (ΔT) = T2 - T1 = 333 K - 283 K = 50 K

Initial volume (V1) = a^3 = (0.1 m)^3 = 0.001 m^3

Change in volume (ΔV) = β × V1 × ΔT
= 3.6 × 10^(-5) K^(-1) × 0.001 m^3 × 50 K
= 1.8 × 10^(-3) m^3
= 1.8 cm^3 (converting to cubic centimeters)

*Answer:* (c) 1.8 cm^3

*Definition of Upper and Lower Fixed Points in Thermometry*

In thermometry, fixed points are reference temperatures used to calibrate thermometers.

- *Upper Fixed Point (UFP):* The temperature at which steam condenses to liquid water at standard atmospheric pressure, defined as 100°C (373.15 K).
- *Lower Fixed Point (LFP):* The temperature at which ice melts to liquid water at standard atmospheric pressure, defined as 0°C (273.15 K).

These fixed points serve as anchors for temperature scales, allowing precise calibration and measurement of temperature

NDSMELODY:
This is why many students fail exams, when you do not understand questions, how will you provide correct solution to such question. The question was not based on linear expansivity, therefore how would you use linear expansion formula......just smiling, you need to do your research very well next time before commenting and in case you are a physics teacher/tutor, kindly consult your text before you teach your students non- sense !!

It clearly stated "Linear expansivity" in the question.

Alusiizizi:
It clearly stated "Linear expansivity" in the question.

bro, the question is volume expansivity not linear expansivity. You should try and understand the question na.

NDSMELODY:
bro, the question is volume expansivity not linear expansivity. You should try and understand the question na.

No point arguing with a stone wall! Bye!

Alusiizizi:
No point arguing with a stone wall! Bye!

okay bye !

Alusiizizi:
No point arguing with a stone wall! Bye!

you are guilty of what you are accusing others of. You need to convert the linear expansivity to volume expansivity.

And the correct relationship is multiplying the linear expansivity by 3.

Question is asking about change in volume not length.

Before you start arguing again, remember a 21st century illiterate is someone who refuses to unlearn and relearn.

chinchum:
you are guilty of what you are accusing others of. You need to convert the linear expansivity to volume expansivity.

And the correct relationship is multiplying the linear expansivity by 3.

Question is asking about change in volume not length.

Before you start arguing again, remember a 21st century illiterate is someone who refuses to unlearn and relearn.

I think he haven't consulted his Text before commenting. I only pity his students .