Question

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)

Answer 1

Answer:

$\large \boxed{\text{28.5 g}}$

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

$\begin{array}{ccl}\text{Heat gained by steel rod + heat lost by water} & = & 0\\m_{1}C_{1} \Delta T_{1} + m_{2}C_{2} \Delta T_{2}& = & 0\\m_{1} \times 0.452 \text{ J ^{\circ} C ^{-1} g ^{-1} }\times 19.50 \, ^{\circ}\text{C} + \text{120 g} \times 4.18\text{ J ^{\circ} C ^{-1} g ^{-1} } \times (-0.50)\, ^{\circ}\text{C}& = & 0\\8.814m_{1}\text{ g}^{-1} - 250.8 & = &0\\\end{array}$

$\begin{array}{ccl}8.814m_{1}\text{ g}^{-1} & = &250.8\\m_{1} & = & \dfrac{250.8}{\text{8.814 g}^{-1}}\\\\& = & \textbf{28.5 g}\\\end{array}\\\text{The mass of the steel rod is \large \boxed{\textbf{28.5 g}} }$