Part A A volume of 115 mL of H2O is initially at room temperature (22.00 C). A c

Question

Part A

A volume of 115 mL of H2O is initially at room temperature (22.00 C). A chilled steel rod at 2.00 C is placed in the water. If the final temperature of the system is 21.50 C , what is the mass of the steel bar? Use the following values: specific heat of water = 4.18 J/(gC) specific heat of steel = 0.452 J/(gC) Express your answer to three significant figures and include the appropriate units.

Part B

The specific heat of water is 4.18 J/(gC). Calculate the molar heat capacity of water. Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

Part ( A)

heat gained by the rod = heat lost by the water

m x S x deltaT = m x S x deltaT

m x 0.452 x ( 21.5 - 2.00 ) = 115 x 4.18 x ( 22.00 - 21.50 )

=> m = 27.27 g

Part (B) :

molar heat capacity = specific heatcapacity x molar mass

molar heat capacity = 4.18 x 18.00 = 75.2 J/(mol.C)

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