Question

The thermochemical equation for the combustion of ethylene gas, $C_2H_4$, is

$C_2H_4(g) + 3O_2(g) \longrightarrow 2CO_2(g) + 2H_2O(l);$

$\Delta H^\Theta = -337 \text{ kcal}$

Assuming 70% efficiency, calculate the weight of water at $20^\circ C$ that can be converted into steam at $100^\circ C$ by burning $1 m^3$ of $C_2H_4$ gas measured at STP. The heat of vaporisation of water at $20^\circ C$ and $100^\circ C$ are $1.00 \text{ kcal kg}^{-1}$ and $540 \text{ kcal kg}^{-1}$ respectively.

Answer 1

170 kg

Answer 2

The problem asks to calculate the mass of water that can be converted into steam by burning a given volume of ethylene gas, considering the efficiency of the process and the heat required for water's phase change and temperature increase.

1. Calculate the total heat released by the combustion of ethylene: The thermochemical equation for the combustion of ethylene is: $C_2H_4(g) + 3O_2(g) \longrightarrow 2CO_2(g) + 2H_2O(l); \Delta H^\Theta = -337 \text{ kcal}$ This means that 1 mole of $C_2H_4$ releases 337 kcal of heat upon combustion.

First, determine the number of moles of $C_2H_4$ in $1 m^3$ at STP. At STP (Standard Temperature and Pressure, $0^\circ C$ and 1 atm), 1 mole of any ideal gas occupies 22.4 L. Given volume of $C_2H_4 = 1 m^3 = 1000 \text{ L}$. Number of moles of $C_2H_4 = \frac{\text{Volume of } C_2H_4}{\text{Molar volume at STP}} = \frac{1000 \text{ L}}{22.4 \text{ L/mol}} \approx 44.643 \text{ mol}$.

Total heat released ($Q_{total}$) by burning $1 m^3$ of $C_2H_4$: $Q_{total} = \text{Moles of } C_2H_4 \times \text{Heat released per mole}$ $Q_{total} = 44.643 \text{ mol} \times 337 \text{ kcal/mol} = 15049.251 \text{ kcal}$.

2. Calculate the useful heat available: The efficiency of the process is given as 70%. Useful heat ($Q_{useful}$) = $Q_{total} \times \text{Efficiency}$ $Q_{useful} = 15049.251 \text{ kcal} \times 0.70 = 10534.4757 \text{ kcal}$.

3. Calculate the heat required to convert water at $20^\circ C$ to steam at $100^\circ C$ per kg of water: This process involves two steps: a. Heating water from $20^\circ C$ to $100^\circ C$: The problem states "heat of vaporisation of water at $20^\circ C$ and $100^\circ C$ are $1.00 \text{ kcal kg}^{-1}$ and $540 \text{ kcal kg}^{-1}$ respectively." The value $1.00 \text{ kcal kg}^{-1}$ for heat of vaporization at $20^\circ C$ is physically incorrect (actual value is around $585 \text{ kcal kg}^{-1}$). It is highly probable that $1.00 \text{ kcal kg}^{-1}$ is intended to be the specific heat capacity of water (c = 1.00 \text{ kcal kg}^{-1} ^\circ C^{-1}), which is a common approximation. We will proceed with this interpretation. Heat required ($Q_1$) = mass ($m$) $\times$ specific heat capacity ($c$) $\times$ temperature change ($\Delta T$) For 1 kg of water: Q_1 = 1 \text{ kg} \times 1.00 \text{ kcal kg}^{-1} ^\circ C^{-1} \times (100^\circ C - 20^\circ C) $Q_1 = 1 \times 1 \times 80 = 80 \text{ kcal}$.

b. Vaporizing water at $100^\circ C$: Heat required ($Q_2$) = mass ($m$) $\times$ latent heat of vaporization ($L_v$) The latent heat of vaporization of water at $100^\circ C$ is given as $540 \text{ kcal kg}^{-1}$. For 1 kg of water: $Q_2 = 1 \text{ kg} \times 540 \text{ kcal kg}^{-1} = 540 \text{ kcal}$.

Total heat required per kg of water ($Q_{per\_kg}$) = $Q_1 + Q_2 = 80 \text{ kcal} + 540 \text{ kcal} = 620 \text{ kcal/kg}$.

4. Calculate the weight (mass) of water that can be converted: Let $m_{water}$ be the mass of water in kg. The useful heat available must be equal to the total heat required to convert $m_{water}$ kg of water. $Q_{useful} = m_{water} \times Q_{per\_kg}$ $10534.4757 \text{ kcal} = m_{water} \times 620 \text{ kcal/kg}$ $m_{water} = \frac{10534.4757 \text{ kcal}}{620 \text{ kcal/kg}} \approx 169.911 \text{ kg}$.

Rounding to a practical number of significant figures (e.g., three), the mass of water is approximately 170 kg.

The final answer is $\boxed{\text{170 kg}}$.

Explanation of the solution:

1. Calculate moles of $C_2H_4$ from its volume at STP ($1 m^3 = 1000 \text{ L}$), using molar volume 22.4 L/mol.
2. Multiply moles of $C_2H_4$ by its molar heat of combustion (337 kcal/mol) to get total heat released.
3. Apply the 70% efficiency to find the useful heat available.
4. Calculate the heat required to convert 1 kg of water from $20^\circ C$ to steam at $100^\circ C$. This involves heating the water to $100^\circ C$ (using specific heat capacity, assumed 1 \text{ kcal kg}^{-1} ^\circ C^{-1}) and then vaporizing it at $100^\circ C$ (using latent heat of vaporization $540 \text{ kcal kg}^{-1}$).
5. Divide the useful heat available by the heat required per kg of water to find the total mass of water that can be converted.