1)A volume of 105 mL of H2O is initially at room temperature (22.00 C). A chille
ID: 475415 • Letter: 1
Question
1)A volume of 105 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 21.40 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)
Express your answer to three significant figures and include the appropriate units.
2) The specific heat of water is 4.18 J/(gC). Calculate the molar heat capacity of water.
Express your answer to three significant figures and include the appropriate units.
Explanation / Answer
1.The temperature difference 22- 21.40 =0.6
The heat release is =105 g × 4.18 J/g × 0.6 =- 263.34 J
The heat release of water = The heat absorption by iron rod
263.34J = m × T × 0.452J/g
Where m = mass of iron rod
T = temperature difference of rod = 21.4- 2= 19.6
m = 263.34J/( 19.4× 0.452J/g) = 30.03g
2.Heat capacity of water = 4.18 J/g
molar mass of water = 18 g /mol
Therefore , molar heat capacity of water = 18 × 4.18J/g
= 75.24 J/mol
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