Part A: A volume of 100. mL of H2O is initially at room temperature (22.00 C). A

Question

Part A: A volume of 100. 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 21.20  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)

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

Please give me the right answer being tired of getting the wrong answers three times of posting question!!

Explanation / Answer

PART A:

Heat lost by the water = heat gained by the steel rod

Q = mcT Q is the quantity of heat, m = mass in g, c is specific heat in J g^-1 ^-deg (kJ/kg K), and T is the temp change.
Heat lost by water = 100.0×4.18×(22.0-21.2) = 334.4 J
Heat gained by the metal = mcT = m×0.452×(21.2- 2.00) = 8.678m
Hence 8.678m = 334.4
m = 38.5 g

PART B

c(H2O) = 4.18 J g^-1 ^-deg MWt H2O = 18.0 hence molar specific heat of H2O is
= 4.18×18.0 = 75.2 J mol^-1 ^-deg

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects