# 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.

How much will the coffee be cooled by adding 50 gm of water at 0 °C?
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
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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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