Calculate the energy in BTUs that is removed

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Homework Help Overview

The problem involves calculating the energy in BTUs required to cool a ton of poultry from an initial temperature of 20 degrees Celsius to a storage temperature of 0 degrees Fahrenheit. The discussion includes considerations of specific heats and latent heat relevant to the cooling process.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the equations for specific heat and latent heat, questioning their applicability given the provided values. There is confusion regarding unit consistency and the conversion of temperatures between Celsius and Fahrenheit.

Discussion Status

Some participants have provided clarifications regarding unit conversions and the importance of maintaining a consistent system of units. There is ongoing exploration of how to apply the equations correctly, but no consensus has been reached on the final approach or calculations.

Contextual Notes

Participants note discrepancies in the mass of the poultry, with conflicting values for one ton. There is also mention of needing to convert between BTUs and kW, which introduces additional considerations for the calculations.

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[SOLVED] Heat problem

Homework Statement


The manager of a fried chicken place wishes to store a ton of poultry at 0 degrees Fahrenheit. Initially the poultry is at 20 degrees Celsius. Calculate the energy in BTUs that is removed in this cooling for storage process.

Assume that the poultry freezes at 32F

Given:
Latent heat: 99 BTU/lb
Specific heats: 0.80 BTU/deg F/lb (above freezing)
0.41 BTU/deg F/lb (below freezing)

Homework Equations


Was not specifically given any but from the book I'm thinking...

Specific heat
Q = cm[tex]\Delta[/tex]T

Latent heat
L = [tex]\frac{|Q|}{m}[/tex]


The Attempt at a Solution


Would I even have to use anyone of those equations? I am already given the specific heat and latent heat so those should do me no good right?

[tex]\Delta[/tex]T = 20 - 0 = 20

Q = (.8)(2000)(20)
Q = 32000J

I know I should have to convert all fahrenheit to celsius.

I am really at a loss though what I can use to give me a final BTU. If anyone could give me a push I'd appreciate it. I have no other sources but Internet and the book so other books will not help at this point. Thanks.
 
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jason.frost said:
[tex]\Delta[/tex]T = 20 - 0 = 20

Q = (.8)(2000)(20)
Q = 32000J

I know I should have to convert all fahrenheit to celsius.
[

Then why don't you? Stick to some consistent set of units.

In the eqn Q = (.8)(2000)(20), where is the 2000 coming from? The 0.8 is in BTU/lb/F, but delta_t = 20...This is meaningless!

Change everything to one system of units.
 
Whoops...sorry I meant to put in what each letter stood for.

Q = cm[tex]\Delta[/tex]T
[tex]\Delta[/tex]T - change in temp
m - mass
c - specific heat (which is given)

1 ton = 2000 lbs = 907.184 kg
kW = Btu / 3414

Q = (.8 BTU/lb/F)(907.184kg)(20C)
Q =

Can I convert the BTUs to kW just by using the equation kW = BTU/3414?
 
1 ton = 2240 lb = 907 kg.

As I said, convert everything to one system of units. Sp heat is given in BTU. Temp change in C can be easily converted to F. Use whichever you like.

> Can I convert the BTUs to kW just by using the equation kW = BTU/3414?

Yes.
 

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