Answer:
1. m₁C₁ΔT₁ + m₂C₂ΔT₂ = 0
2. 207 g
Explanation:
This is a problem in calorimetry — the measurement of the quantities of heat that flow from one object to another.
It is based on the Law of Conservation of Energy — Energy can be transformed from one type to another, but it cannot be destroyed or created.
If heat flows out of one object, the same amount of heat must flow into another object.
Since there is no change in total energy,
Heat₁ + heat₂ + heat₃+ … = 0
The symbol for the quantity of heat transferred is q, so we can rewrite the word equation as
q₁ + q₂ + q₃ + … = 0
The formula for the heat absorbed or released by an object is
q = mCΔT, where
m = the mass of the sample
C = the specific heat capacity of the sample, and
ΔT = T_f - T_i = the change in temperature
1. Equation
There are two heat flows in this problem,
Heat from steel rod + heat absorbed by water = 0
q₁ + q₂ = 0
m₁C₁ΔT₁ + m₂C₂ΔT₂ = 0
2. Data:
m₁ = ?; C₁ = 0.452 J°C⁻¹g⁻¹;T_f = 21.20 °C; T_i = 2.50 °C
V₂ = 150.0 mL; C₂ = 4.184 J°C⁻¹g⁻¹; T_f = 21.20 °C; T_i = 24.00 °C
3. Calculations
(a) m₂
The density of water is 0.9982 g/mL, so
m₂ = 150.0 mL × 0.9982 g/mL = 149.7 g
(b) Temperature changes
[tex]\Delta T_{1} = T_{\text{f}} - T_{\text{i}} = 21.20 \, ^{\circ}\text{C} - 2.50 \, ^{\circ}\text{C} = 18.70 \, ^{\circ}\text{C}\\\Delta T_{2} = T_{\text{f}} - T_{\text{i}} = 21.20 \, ^{\circ}\text{C} - 24.00 \, ^{\circ}\text{C} = -2.80 \, ^{\circ}\text{C}[/tex]
(c) q₁
q₁ = m₁C₁ΔT₁ = m₁ × 0.452 J°C⁻¹g⁻¹ × 18.70 °C = 8.452m₁ J·g⁻¹
(d) q₂
q₂ = m₂C₂ΔT₂ = 149.7 g × 4.184 J°C⁻¹g⁻¹ × (-2.80 °C) = -1754 J
(e) m₁
[tex]\begin{array}{rcl}q_{1} + q_{2} & = & 0\\8.452m_{1} \, \text{J}\cdot\text{g}^{-1} + (-\text{1754 J}) & = & 0\\8.452m_{1}\,\text{g}^{-1} - 1764 &=& 0\\m_{1} & = & \dfrac{1764}{8.452 \,\text{g}^{-1}}\\\\ & = & \textbf{207 g}\\\end{array}\\\text{ The mass of the steel rod is $\large \boxed{\textbf{207 g}}$}[/tex]
