Cooling a Cup of Coffee

You have a 200 gram cup of coffee at 100 °C, too hot to drink. Various cooling strategies demonstrate specific heat, phase changes, and the approach to thermal equilibrium.

Add cold water
How much will the coffee be cooled by adding 50 gm of water at 0 °C?
Add ice
How much will the coffee be cooled by adding 50 gm of ice at 0 °C?
Force evaporation
Suppose you start with 300 gm of coffee at 100 °C and force 50 grams of it to vaporize, leaving the final mass at 250 gm?
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Cooling a Cup of Coffee

You have a 200 gram cup of coffee at 100 °C, too hot to drink. How much will you cool it by adding 50 gm of water at 0 °C?

Heat lost by coffee = Heat gained by water
-Qcoffee = Qwater
-cmcΔTcoffee = cmwΔTwater


20,000 - 200Tf = 50Tf
Cool with ice
Cool by vaporization
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Cooling a Cup of Coffee

You have a 200 gram cup of coffee at 100 °C, too hot to drink. How much will you cool it by adding 50 gm of ice at 0 C?

Heat lost by coffee = Heat gained by ice
-Qcoffee = Qice
-cmcΔTcoffee = miLf + cmiΔTice


20,000cal - 200Tf = 4000cal + 50Tf
Vary parameters
Cool with water
Cool by vaporization
 Index

Heat transfer concepts

Heat transfer examples
 
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Cooling a Cup of Coffee

You have a 300 gram cup of coffee at 100 °C, too hot to drink. How much will you cool it by forcing 50 gm to evaporate, leaving 250 gm?

Heat lost by coffee = Heat of vaporization
-Qcoffee = Qvaporization
-cmcΔTcoffee = mvLv

(1 cal/gm°C)(250gm)(100-Tf)=(540cal/gm)(50gm)

25,000 - Tf = 27,000

But this can't be right because it gives a negative temperature (-8°C) and the specific heat equation Q = cmΔT is valid only so long as a phase change is not encountered, so we can't pass 0°C with this equation. If 25000 calories are extracted, we have cooled the coffee to 0°C but still have 2000 cal to remove. This will freeze some of the coffee:

(2000 cal)/(80 cal/gm) = 25 gm frozen at 0°C
Cool with water
Cool with ice
 Index

Heat transfer concepts

Heat transfer examples
 
HyperPhysics***** Thermodynamics R Nave
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