Question

Part a. A volume of 90.0 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 21.10  ∘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

a)

m(rod) = to be calculated

C(rod) = 0.452 J/goC

m(water) = 90.0 g

T(water) = 22.0 oC

C(water) = 4.18 J/goC

T = 21.1 oC

We will be using heat conservation equation

use:

heat lost by water = heat gained by rod

m(water)*C(water)*(T(water)-T) = m(rod)*C(rod)*(T-T(rod))

90.0*4.18*(22.0-21.1) = m(rod)*0.452*(21.1-2.0)

m(rod)= 39.2 g

Answer: 39.2 g

b)

Molar mass of H2O,

MM = 2*MM(H) + 1*MM(O)

= 2*1.008 + 1*16.0

= 18.016 g/mol

use:

molar heat capacity = specific heat capacity * molar mass

molar heat capacity = 4.18 J/g.oC * 18.016 g/mol

molar heat capacity = 75.31 J/mol.oC

Answer: 75.3 J/mol.oC