calculate the amount of heat needed to boil 116. g of ethanol (ch₃ch₂oh), beginning from a temperature of −75.3 °c. round your answer to 4 significant digits. also, be sure your answer contains a unit symbol.

Question

Accepted Answer

To solve the problem of calculating the heat needed to boil ethanol, we need to consider three steps: heating the ethanol from $-75.3^\circ \text{C}$ to its boiling point, vaporizing the ethanol at its boiling point. First, we need the specific heat capacity of liquid ethanol ($c_{\text{liquid}}$), the enthalpy of vaporization ($\Delta H_{\text{vap}}$), and the boiling point of ethanol ($T_b = 78.4^\circ \text{C}$).

### Step 1: Calculate the moles of ethanol
The molar mass of ethanol ($\text{CH}_3\text{CH}_2\text{OH}$) is:
$$
\begin{align*}
M &= 2(12.01) + 6(1.008) + 16.00 \
&= 24.02 + 6.048 + 16.00 \
&= 46.068 \, \text{g/mol}
\end{align*}
$$
Moles of ethanol ($n$):
$$
n = \frac{m}{M} = \frac{116 \, \text{g}}{46.068 \, \text{g/mol}} \approx 2.518 \, \text{mol}
$$

### Step 2: Heat the liquid ethanol from $-75.3^\circ \text{C}$ to $78.4^\circ \text{C}$
The specific heat capacity of liquid ethanol ($c_{\text{liquid}}$) is $2.46 \, \text{J/g}^\circ\text{C}$.
The temperature change ($\Delta T$):
$$
\Delta T = 78.4^\circ \text{C} - (-75.3^\circ \text{C}) = 153.7^\circ \text{C}
$$
Heat required for heating ($q_1$):
$$
q_1 = mc\Delta T = 116 \, \text{g} \times 2.46 \, \text{J/g}^\circ\text{C} \times 153.7^\circ \text{C}
$$
$$
q_1 \approx 116 \times 2.46 \times 153.7 \approx 44700 \, \text{J} = 44.7 \, \text{kJ}
$$

### Step 3: Vaporize the ethanol at its boiling point
The enthalpy of vaporization of ethanol ($\Delta H_{\text{vap}}$) is $38.56 \, \text{kJ/mol}$.
Heat required for vaporization ($q_2$):
$$
q_2 = n\Delta H_{\text{vap}} = 2.518 \, \text{mol} \times 38.56 \, \text{kJ/mol} \approx 97.1 \, \text{kJ}
$$

### Step 4: Total heat required
$$
q_{\text{total}} = q_1 + q_2 = 44.7 \, \text{kJ} + 97.1 \, \text{kJ} = 141.8 \, \text{kJ}
$$

Wait, let's check the calculations again for accuracy.

### Recalculating Step 2:
$$
q_1 = 116 \, \text{g} \times 2.46 \, \text{J/g}^\circ\text{C} \times 153.7^\circ \text{C}
$$
$$
116 \times 2.46 = 285.36
$$
$$
285.36 \times 153.7 \approx 285.36 \times 150 + 285.36 \times 3.7 = 42804 + 1055.832 = 43859.832 \, \text{J} \approx 43.86 \, \text{kJ}
$$

### Recalculating Step 3:
$$
q_2 = 2.518 \, \text{mol} \times 38.56 \, \text{kJ/mol}
$$
$$
2.518 \times 38.56 \approx 2.5 \times 38.56 + 0.018 \times 38.56 = 96.4 + 0.694 = 97.094 \, \text{kJ}
$$

### Total heat:
$$
q_{\text{total}} = 43.86 \, \text{kJ} + 97.094 \, \text{kJ} \approx 140.95 \, \text{kJ} \approx 141.0 \, \text{kJ} \, (\text{to 4 significant figures})
$$

Wait, maybe I made a mistake in the specific heat capacity or enthalpy of vaporization. Let's confirm the values:

- Specific heat capacity of liquid ethanol: $2.46 \, \text{J/g}^\circ\text{C}$ (correct)
- Boiling point of ethanol: $78.4^\circ \text{C}$ (correct)
- Enthalpy of vaporization of ethanol: $38.56 \, \text{kJ/mol}$ (correct)

Moles of ethanol:
$$
n = \frac{116 \, \text{g}}{46.07 \, \text{g/mol}} \approx 2.518 \, \text{mol} \, (\text{correct})
$$

Temperature change:
$$
\Delta T = 78.4 - (-75.3) = 153.7^\circ \text{C} \, (\text{correct})
$$

Heat for heating:
$$
q_1 = 116 \times 2.46 \times 153.7 = 116 \times 378.102 = 43859.832 \, \text{J} = 43.86 \, \text{kJ} \, (\text{correct})
$$

Heat for vaporization:
$$
q_2 = 2.518 \times 38.56 = 97.09 \, \text{kJ} \, (\text{correct})
$$

Total heat:
$$
43.86 + 97.09 = 140.95 \, \text{kJ} \approx 141.0 \, \text{kJ}
$$

So the total heat required is approximately $\boldsymbol{141.0 \, \text{kJ}}$ (to 4 significant figures).

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QUESTION IMAGE

Question

Explanation:

Step 1: Calculate the moles of ethanol

Step 2: Heat the liquid ethanol from \(-75.3^\circ \text{C}\) to \(78.4^\circ \text{C}\)

Step 3: Vaporize the ethanol at its boiling point

Step 4: Total heat required

Recalculating Step 2:

Recalculating Step 3:

Total heat:

Answer: