A 10 g object at 50oC is placed in 100 g water at 39.8oC in a...

A 10 g object at 50^oC is placed in 100 g water at 39.8^oC in a calorimeter. The temperature of the water and the object equilibrated at 40.0^oC. Identify the metal.

Please explain this. Thanks!

Expert Solution

heat lose of metal = heat gain of water

mct = mct

10*c*(50-40) = 100*4.184*(40-39.8)

c = 0.84J/g-c⁰

metal specific heat = 0.84J/g-c⁰ . It is aluminium metal

queen_honey_blossom answered 2 years ago

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