A 53-g piece of an unknown metal is heated to 119°c and then placed into a calorimeter containing 165g of water at a temperature of 19.0°c.The final temperature of both is 32.0°c. What is the specific heat of the unknown metal?

Question

1286 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-13T16:46:25

These problems require considering the heat source and the water as its own equilibrium. Recall,

#q = mC_sDeltaT#

The heated metal will lose heat, and the water will gain heat.

So,

#-q_"M" = q_(H_2O)#

and by extension,

#-m_"M"C_"M"DeltaT_"M" = m_(H_2O)C_(H_2O)DeltaT_(H_2O)#

The problem states #T_"f"# for both is #32.0°"C"# once the system has equilibrated.

Hence,

#=> C_"M" = (-m_(H_2O)C_(H_2O)DeltaT_(H_2O))/(m_"M"DeltaT_"M") approx (1.94"J")/(g*°"C")#

is the specific heat of the metal given your data.