Part A A volume of 100. mL of H2O is initially at room temperature (22.00 C). A

Question

Part A

A volume of 100. 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.

please help asap

Explanation / Answer

100 x 4.18 x (22.0-21.5 ) = 0.452 x mass of steel bar x (21.5-2)

mass of steel bar = 23.71 gm

Part B

molar heat capacity is J/mole C not J/gC
4.18 J/gC x (18.0 g / mole) = 75.2 J/moleC

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