A volume of 125 mL of H2O is initially at room temperature 2

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 21.40 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.

Solution

Heat gained by chilled steel rod = heat gained by water

Heat =mass x sp heat x (T₂ -T₁)

Mass of water = v xd = 125 ml x 1g/ml = 125 g(density of water = 1g/ml)

Heat lost by water (-q) =125 g x 4.18 j/g⁰C x (21.4 - 22)⁰C (-ve sign indicates heat release)

Heat gained by steel rod (q)= m x 0.452 J/g⁰C x (21.4- 2)⁰C

ie m x 8.7688 = 313.5g

m = 313.5 g /8.7688 = 35.75g  

Mass of steel rod = 35.8 g

Molar heat capacity is the heat needed to raise the temperature of 1 mol (ie 18 g ) of water through 1 degree C.

Sp heat (ie the heat needed to raise the temperature of 1 g of water through 1 degree C) of water = 4.18 J/g⁰C

Molar heat capacity = 4.18 J/g⁰C x 18 g/mol = 75.24 J/mol

Molar heat capacity = 75.24 J/mol