A volume of 85.0 mL of H2O is initially at room temperature (22.00 C). A chilled

Question

A volume of 85.0 mL of H2O is initially at room temperature (22.00 C). A chilled steel rod at 2.00 Cis 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.

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

water is lossing heat and steel rod is abosorbinf heat

so lets make the set up as follows

- q water = + q steel rod

-m*c*delta T = m*x*delta T

lets put the values in the formula

-85 g * 4.18 J per g C * (21.3 C - 22.00 C) = m * 0.452 J per g C * (21.3 C - 2.00 C)

248.71 = 8.7236 * m

248.71 / 8.7236 = m

28.5 g = m

so mass of steel rod = 28.5 g

part 2

molar specific heat of water = specific heat of water * molar mass of water

                                       = 4.18 J per g C * 18.014 g per mol

                                      = 75.3 J per mol C

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