Question

Part A A volume of 115 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. View Available Hint(s) mass of the steel - Value Units Submit

Answer 1

A)
m(rod) = to be calculated

C(rod) = 0.452 J/goC
Since density of water is 1 g/mL and volume is 115 mL,
m(H2O) = 115 g
T(H2O) = 22.0 oC
C(H2O) = 4.184 J/goC
T = 21.1 oC
We will be using heat conservation equation

use:
heat lost by H2O = heat gained by rod
m(H2O)*C(H2O)*(T(H2O)-T) = m(rod)*C(rod)*(T-T(rod))
115.0*4.184*(22.0-21.1) = m(rod)*0.452*(21.1-2.0)
m(rod)= 50.1603 g
Answer: 50.16 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