1) A 0.0545 kg chunk of an unknown metal that has been in boiling water for seve

Question

1) A 0.0545 kg chunk of an unknown metal that has been in boiling water for several minutes is quickly dropped into an insulating StyrofoamTM beaker that contains 0.198 kg of water at room temperature (20.0 C). After waiting for a few minutes, you observe that the water's temperature has reached a constant value of 22.0 C

Part A- Assuming that the StyrofoamTM absorbs a negligibly small amount of heat and that no heat was lost to the surroundings, what is the specific heat of the metal? in joules per kilogram-kelvin

Explanation / Answer

Since the metal was in boiling water for some time we can assume its temperature is 100oC

Now Qout of the metal = Qinto Water (Heat out = Heat in)

Use Q = m*c*deltaT

So (metal) 0.0545*c*(100 - 22.0) = 0.198*4190*(22.0 - 20.0) = 1660
So c = 1660/(.545*(100-22.5)) = 40/kg-oC

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