In: Chemistry
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)
Heat in thermodynamics is a form of energy which is either absorbed or released from the reaction mixture or any thermodynamic process.
The relation between heat of the system and the temperature is:
where,
Q is the heat absorbed or released.(Heat loss is shown by positive sign and heat gain is shown by negative sign)
m is the mass
T is the change in temperature
c is the specific heat. It is defined as heat added to 1 unit mass of the sample to raise its temperature by 1 0Celsius.
Given,
Volume of water = 80 mL
Room temperature or Initial temperature of water = 22.0 0C
Final temperature of water = 21.5 0C
Specific heat of water = 4.18 J/g0C
The change in the temperature is :
Delta T = 22.0 0C - 21.5 0C = 0.5 0C
Considering the density of water = 1 g/mL
So the mass of water is
mass of water = 80 g
Heat lost by water, Q:
Heat lost by water is heat gained by steel
so,
Negative sign signifies that the heat is gained.
Initial temperature of the steel bar = 2.0 0C
Final temperature of steel bar = 21.5 0C
Specific heat of steel bar = 0.452 J/g0C
The change in the temperature is :
Delta T = 2.0 0C - 21.5 0C = -19.5 0C
So applying the formula of heat gained by the steel bar as:
Mass of the steel bar:
So, the mass of the steel bar is 18.97 g