Question

Part A) A volume of 80.0 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/(g⋅∘C)

specific heat of steel = 0.452 J/(g⋅∘C)

Express your answer to three significant figures and include the appropriate units.

Part B) The specific heat of water is 4.18 J/(g⋅∘C). Calculate the molar heat capacity of water.
Express your answer to three significant figures and include the appropriate units.

Answer 1

Part1)Calculate:

ms= - 4.18 J/(g∘C) * (21.10 G∘C – 22.00 ∘C) * 110.00 g / (0.452 J/(g∘

C) /* (21.10 ∘C –2.00 ∘C))= 47.93 g

Part2)

4.18 x 18 =75,24.