Answer:
[tex]\large \boxed{\text{933 J}}[/tex]
Explanation:
There are three heat transfers involved.
heat from combustion of propane + heat gained by water + heat gained by calorimeter = 0
q₁ + q₂ + q₃ = 0
m₁ΔH + m₂C₂ΔT + C_calΔT = 0
Data:
m₁ = 2.1 g
m₂ = 280 g
Ti = 25.00 °C
T_f = 26.55 °C
Ccal = 92.3 J·°C⁻¹
Calculations:
Let's calculate the heats separately.
1. q₁
q₁ = 2.1 g × ΔH = 2.1ΔH g
2. q₂
ΔT = T_f - Ti = 26.55 °C - 25.00 °C = 1.55 °C
q₂ = 280 g × 4.184 J·°C⁻¹ × 1.55 °C = 1816 J
3. q₃
q₃ = 92.3 J·°C⁻¹ × 1.55 °C = 143.1 J
4. ΔH
[tex]\begin{array}{rcl}\text{2.1$\Delta$H g +1816 J +143.1 J} & = & 0\\\text{2.1$\Delta$H g +1959 J} & = & 0\\\text{2.1$\Delta$H g}& = & \text{-1959 J}\\\Delta H & = & \dfrac{\text{-1959 J}}{\text{2.1 g}}\\\\& = & \textbf{-933 J/g}\\\end{array}\\\text{The combustion releases $\large \boxed{\textbf{933 J}}$ per gram of fuel burned.}[/tex]
