A piece of metal of mass 35.0 g at 100.0°C was placed in 150.0 g of...

Question

Question

A piece of metal of mass 35.0 g at 100.0°C was placed in 150.0 g of water at 20.0 °C. After stirring, the final temperature o

Answer 1

Answer #1

Heat absorbed by the water = Heat loosed by Metal

M(water)*C*(Tf-Ti(water)) = M(metal)*C(metal)*(Ti(metal)-Tf)

150*4.184*(23.8-20) = 35*C(metal)*(100-23.8)

C(metal) = 150*4.184*(23.8-20)/(35*(100-23.8))

Cmetal = 0.89 J/g-C

Answer 2

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