A volume of 120 mL of H2O is initially at room temperature (2200 degree C). A ch
ID: 1324543 • Letter: A
Question
A volume of 120 mL of H2O is initially at room temperature (2200 degree C). A chilled steel rod at 2.00 degree c is placed in the water. If the final temperature of the system is 21.10 degree C, what is the mass of the steel bar? Use the following values: specific heat of water = 4.18 J/(g. degree C) specific heat of steel = 0.452 J/(g. degree C) Express your answer to three significant figures and include the appropriate units. mass of the steel = The specific heat of water is 4.18 J/(g. degree C). Calculate the molar heat capacity of water Express your answer to three significant figures and include the appropriate units. molar heat capacity for waterExplanation / Answer
120 mL of water = 120 g of water
As
m1c1(Tf1 - Ti1) = m2c2(Ti2 - Tf2)
where material 1 is water, material 2 is steel, then, solving for m2,
m2 = [m1c1(Tf1 - Ti1)]/[c2(Ti2 - Tf2)]
m2 = 52.3 g [ANSWER]
**************
Since H20 = 18 g/mol, then
4.18 J/gC * 18 g/mol = 75.2 J/[mol K] [ANSWER]
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