Thermodyanmics, thermal properties of matter

Click For Summary
The discussion revolves around calculating the minimum energy required to freeze 200mL of ethyl alcohol at 20 degrees Celsius. The melting point was initially misidentified as 114 degrees, but it was clarified to be -114 degrees. The calculations for energy removal included both the specific heat and the latent heat of fusion, but the temperature change was incorrectly calculated. The correct temperature change should be from 20 to -114 degrees, significantly affecting the total energy calculation. The final consensus is that the initial calculations were incorrect due to the misunderstanding of the temperature range.
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
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K