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

b is the correct

we can conclude that the melting point of table salt is above room temperature (it's actually 801 deg c). the melting point of mercury must be below room temperature (- 39 deg c).