A volume of 120 mL of H2O is initially at room temperature (22.00 ∘C ). A chilled steel rod at 2.00 ∘C ∘C is placed in the water. If the final temperature of the system is 21.50 ∘C ∘C , what is the mass of the steel bar? Use the following values: specific heat of water = 4.18 J/(g⋅∘C)J/(g⋅∘C) specific heat of steel = 0.452 J/(g⋅∘C)

The mass of the steel bar is "28.5 g"

According to the question,

Mass,

Specific heat of steel,

Specific heat of water,

Temperature,

As we know,

→       Heat lost by water = Heat gained by steel

then,

→                            

By substituting the values, we get

→

→                                    

→                                        

→                                        

Thus the above answer is appropriate.

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Explanation:

There are two heat flows in this process and, since energy (heat) can neither be destroyed nor created, the energy change for the system must equal zero.

Data:

For Fe,    m₁ = ?;         C₁ = 0.452 J°C⁻¹g⁻¹; Ti =   2.00 °C; T_f = 21.50 °C

For H₂O, m₂ = 120 g; C₂ = 4.18    J°C⁻¹g⁻¹; Ti = 22.00 °C; T_f = 21.50 °C

Calculations:

1. Temperature changes

ΔT₁ = T_f - Ti = 21.50 °C -   2.00 °C = 19.50 °C

ΔT₂ = T_f - Ti = 21.50 °C - 22.00 °C = -0.50 °C

2. Mass of steel rod

the answer is the