Part A A volume of 125 mL of H2O is initially at room temperature (22.00 C). A c

Question

Part A

A volume of 125 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 21.30  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

Initial volume of H2O = 125 ml

initial temperature of H2O = 22.00 C

initial temperature of steel rod = 2.00 C

Final temperature of the system = 21.30  C

specific heat of water = 4.18 J/(gC)

specific heat of steel = 0.452 J/(gC)

mass of steel rod= To be calculated

the decrease in temperature of water = 295-294.3= 0.7 C or 0.7 kelvin

Amount of energy absorbed by the water is 4.18 * 0.7 * 125 =365.75 J

increase in temperature of steel bar = 21.3-2= 19.3 °C.

Q(energy) = Mass * specfic heat *delta T(change in temperature)

mass =Specfic heat of steel *delta T(change in temperature)/Q(energy)

mass =365.75/ ( 19.3 * 0.452)

mass of steel = 365.75/ 8.7236 grams

mass of steel = 41.9 (or) 42 grams

Part-B

The specific heat of water is 4.18 J/(gC). Calculate the molar heat capacity of water

4.18 J/gC x (18.0 g / mole) = 75.2 J/mole C

the molar heat capacity =75.2 joules/Mole degree C

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