Problem 4.9 - ⋅ = when τ 2 hr ⋅ = Hence k 1 τ ln T...

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Problem 4.9 [Difficulty: 2] Given: Data on cooling of a can of soda in a refrigerator Find: How long it takes to warm up in a room Solution: The First Law of Thermodynamics for the can (either warming or cooling) is Mc dT dt k TT amb () = or dT dt k TT amb () = where M is the can mass, c is the average specific heat of the can and its contents, T is the temperature, and T amb is the ambient temperature Separating variables dT TT amb A dt = Integrating Tt () T amb T init T amb () e A t + = where T init is the initial temperature. The available data from the coolling can now be used to obtain a value for constant A Given data for cooling T init 80 °F = T init 540 R = T amb 35 °F = T amb 495 R = T4 5 ° F = T 505 R
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Unformatted text preview: ⋅ = when τ 2 hr ⋅ = Hence k 1 τ ln T init T amb − T T amb − ⎛ ⎜ ⎝ ⎞ ⎠ ⋅ = 1 2 hr ⋅ 1 hr ⋅ 3600 s ⋅ × ln 540 495 − 505 495 − ⎛ ⎝ ⎞ ⎠ × = k 2.09 10 4 − × s 1 − = Then, for the warming up process T init 45 °F = T init 505 R ⋅ = T amb 72 °F = T amb 532 R ⋅ = T end 60 °F = T end 520 R ⋅ = with T end T amb T init T amb − ( ) e k − τ ⋅ + = Hence the time τ is τ 1 k ln T init T amb − T end T amb − ⎛ ⎜ ⎝ ⎞ ⎠ ⋅ = s 2.09 10 4 − ⋅ ln 505 532 − 520 532 − ⎛ ⎝ ⎞ ⎠ ⋅ = τ 3.88 10 3 × s = τ 1.08 hr ⋅ =...
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This note was uploaded on 10/19/2011 for the course EGN 3353C taught by Professor Lear during the Fall '07 term at University of Florida.

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