Thermodyanmics, thermal properties of matter

Click For Summary

Homework Help Overview

The problem involves calculating the minimum amount of energy that must be removed from a beaker containing 200mL of ethyl alcohol at 20 degrees Celsius to produce solid ethyl alcohol. The relevant properties include the melting/freezing point, latent heat, density, and specific heat capacity of ethyl alcohol.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculations involving mass, specific heat, and latent heat. There is a focus on the temperature change and the interpretation of the melting/freezing point of ethyl alcohol.

Discussion Status

Some participants have identified a potential error in the original poster's understanding of the melting/freezing point, suggesting that the temperature should be -114 degrees Celsius instead of 114 degrees. This has led to a reevaluation of the temperature change used in the calculations.

Contextual Notes

There is a discrepancy in the melting/freezing point of ethyl alcohol, which affects the calculations. The original poster initially used an incorrect value, leading to confusion regarding the energy calculations.

Jennifer001
Messages
22
Reaction score
0

Homework Statement


a beaker contains 200mL of ethyl alcohol at 20degrees. what is the minumum amount of energy that must be removed to produce solid ethyl alcohol..

the melting/freezing pt is 114degrees Lf=1.09x10^5
density is 790kg/m^3
specific heat cap = 2400J/kgK

Homework Equations


Q=mcdT
Q=ML
m=pv

The Attempt at a Solution


m=pv=(2.00*10^-4m^3)(790)=0.158kg
Q1=mcdT
=0.158(2400)(94)
=35644.8J
Q2=ML
=0.158(1.09*10^5)
=17222J

Qtotal=Q1+Q2
=52866.8J

i think i did somethign wrong because in the book it says the answer is 68000J can someone help me out i don't know what i did wrong?
 
Physics news on Phys.org
Jennifer001 said:

Homework Statement


a beaker contains 200mL of ethyl alcohol at 20degrees. what is the minumum amount of energy that must be removed to produce solid ethyl alcohol..

the melting/freezing pt is 114degrees

That 114 degrees must be wrong. If it were true, it would already be a solid at 20 degrees.

Lf=1.09x10^5
density is 790kg/m^3
specific heat cap = 2400J/kgK

Homework Equations


Q=mcdT
Q=ML
m=pv

The Attempt at a Solution


m=pv=(2.00*10^-4m^3)(790)=0.158kg
Q1=mcdT
=0.158(2400)(94)
=35644.8J
Q2=ML
=0.158(1.09*10^5)
=17222J

Qtotal=Q1+Q2
=52866.8J

i think i did somethign wrong because in the book it says the answer is 68000J can someone help me out i don't know what i did wrong?
 
Redbelly98 said:
That 114 degrees must be wrong. If it were true, it would already be a solid at 20 degrees.


oooh sorry that 114 is suppose to be -114
 
Jennifer001 said:
oooh sorry that 114 is suppose to be -114

Okay, then do you see the error here:

Jennifer001 said:
Q1=mcdT
=0.158(2400)(94)
=35644.8J
 
Redbelly98 said:
Okay, then do you see the error here:

oo i added it wrong for dT its suppose to be -114-20 instead of 114-20

ok thank you
 

Similar threads

Energy resultant of Solidification Point Of Ethanol
  • · Replies 1 ·
Replies
1
Views
3K