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Article #1287

microwave cooking

#1287
From: LUDVIK KOWALSKI
Date: Wed, 14 Sep 1994 04:07
22 lines
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>Any Physical science text book from the 9th grade on up will discuss thermal
>energy transfer and specific heat. Obviously water molecules absorb micro wave
>radiation many times more efficiently than metals which should reflect this
>form of 'light' (EM radiation). A calorimeter is used to measure changes in
>temperature and so calculate changes in thermal energy and specific heats.
>Landon

You are right, Landon, that the amount of heat absorbed can always be measured
with a calorimeter. And that's what really counts in cooking. But I want to
PREDICT outcomes and I do not know how. Here is the problem; put yourself in
my place, and try to solve it.

Ten grams of water and ten grams of honey (or oil, blood, meat, wood, etc.)
are heated in a microwave oven under identical conditions. Which substance
will absorb more heat? What is the heat absorption ratio in two substances?

Try to answer before facts are known from direct measurements. I tried but was
not successful. The elementary physics textbooks I used did not prepare me to
deal with microwave cooking in quantitative ways. Knowing specific heats of
two selected substances does not help to solve the problem.

              Ludwik Kowalski   kowalski@apollo.montclair.edu
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Date: Tue, 13 Sep 1994 23:07:00 EST
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From: LUDVIK KOWALSKI <KOWALSKI@APOLLO.MONTCLAIR.EDU>
Subject: microwave cooking
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