Carl C.

asked • 04/21/16

Heat transfer between metal and water, what amount of water is boiled?

You place a very hot piece of aluminum (T=120 degrees C, c=900 J/(kg*degrees C) into a 500ml beaker of water that is already at an initial temperature of 90.0 degrees C. The mass of this piece of aluminum is 0.300kg. What mass of water will boil as a result?

So I know Q=mc(change T), and I know that the problem sets up like this: mc(change T) (for aluminum)=mc(change T) ( for water) + m(boil)L(vaporization).

Plugging the numbers in in the same order: (0.300kg)(900 J/(kg*C))(120C-100C)=(0.500kg)(4186 J/(kg*C))(120C-90C)+m(boil)(225600 J/kg).

Solving for m(boil) equals -0.025.

I understand that you cannot have a negative amount of mass boiling. What does it mean to get a negative number as the answer? What causes this and how could the problem be written differently to avoid this issue?

1 Expert Answer


Steven W. answered • 08/09/16

Physics Ph.D., college instructor (calc- and algebra-based)

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