Question

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)

Answer 1

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