The Physics
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Opus in profectus

Latent Heat

Practice

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practice problem 2

In order to extract the maximum flavor in the shortest amount of time, your local fast food purveyor has decided to brew its coffee at 90 ℃ and serve it quickly so that it has only cooled down to 85 ℃. While this may be economically sensible, it is negligent and dangerous from a health and safety standpoint. Water (which is what coffee mostly is) at 85 ℃ is hot enough to cause third-degree burns (the worst kind) in two to seven seconds. You decide to add ice cubes to your coffee to cool it down to a more reasonable 55 ℃ so you will be able to drink it sooner. (Watery brew be damned. You need your caffeine now.) How many 23.5 g ice cubes at −18.5 ℃ should you add to your 355 ml cup of coffee to accomplish your thermal goal?

solution

This is a conservation of energy problem. The heat gained by the ice will be equal to the heat lost by the coffee.

+Qice = −Qcoffee

This mixing problem is more complicated than the ones in the previous section, however. The ice must first warm up to its melting point (a temperature change), then it has to melt (a phase change), and then the liquid has to warm up (another temperature change). The coffee has less to do. It just has to cool down.

 Qcold ice + Qmelting + Qmelted ice = − Qcoffee
 [mcΔT] cold ice + [mL] melting + [mcΔT] melted ice = − [mcΔT] coffee

The final mixture will end up at one temperature. (Watch the order of subtraction when dealing with temperature changes.)

 mice(2,090 J/kg C°)(0 − −18.5 ℃) + mice(334,000 J/kg) + mice(4,200 J/kg C°)(55 − 0 ℃) = − (0.355 kg)(4,200 J/kg C°)(55 − 85 ℃)
 mice(603,665 J/kg) = (44,730 J) mice = 0.0741 kg

This is about three ice cubes.

number = (74.1 g)/(23.5 g) ≈ 3 ice cubes

practice problem 3

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