An object with a mass of 21 g is dropped into 400 mL of water at 0^@C. If the object cools by 150 ^@C and the water warms by 14 ^@C, what is the specific heat of the material that the object is made of?

1 Answer
MeneerNask
Feb 19, 2016

The heat energy lost by the object is the same as the energy gained by the water, or Q_o=Q_w

Explanation:

c_o*m_o*DeltaT_o=c_w*m_w*DeltaT_w

Now fill in what we know (c_w=4.2 and 1mL=1g water):
c_o*21*150=4.2*400*14

Work this out:
c_o=(4.2*400*14)/(21*150)~~7.47

This is a ridiculously high outcome, as no known material has a specific heat this high. Probably a 0 was lost or added in the problem.
With a mass of 210g or an amount of water of 40mL the outcome would be 0.747, which is more plausible.

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