Question

water is frequently used to generate steam, a product you can sometimes see as you walk down a city street (think steam rising up a street grate). using the heating curve and data below, determine the amount of energy needed (kj) to heat 6.9 g of water from 180.6 °f to 250.0 °f. (enter answer in regular notation with one digit after the decimal place) heating curve graph and data table: ( c_{\text{solid}} = 2.09 , \text{j/g°c} ), ( c_{\text{liquid}} = 4.184 , \text{j/g°c} ), ( c_{\text{gas}} = 2.03 , \text{j/g°c} ), ( delta h_{\text{fus}} = 6.01 , \text{kj/mol} ), ( delta h_{\text{vap}} = 40.6 , \text{kj/mol} )

Explanation:

Step1: Convert Fahrenheit to Celsius

The formula to convert Fahrenheit to Celsius is $^\circ C=\frac{5}{9}(^\circ F - 32)$.
For the initial temperature: $^\circ C_1=\frac{5}{9}(180.6 - 32)=\frac{5}{9}\times148.6\approx82.56^\circ C$
For the final temperature: $^\circ C_2=\frac{5}{9}(250.0 - 32)=\frac{5}{9}\times218\approx121.11^\circ C$

Step2: Determine the phase of water

From the heating curve, at $82.56^\circ C$ water is liquid, and at $121.11^\circ C$ water is gas (steam). But first, we need to heat liquid water to $100^\circ C$, then vaporize it (but wait, no—wait, the temperature range is from ~82.56 to ~121.11. Wait, no, wait: Wait, the problem is heating from 180.6°F (82.56°C) to 250.0°F (121.11°C). Wait, but water boils at 100°C. So first, heat liquid water from 82.56°C to 100°C, then vaporize it (phase change), then heat the steam from 100°C to 121.11°C. Wait, but let's check the data. Wait, the given data: $C_{liquid}=4.184\ J/g^\circ C$, $\Delta H_{vap}=40.6\ kJ/mol$, $C_{gas}=2.03\ J/g^\circ C$.

First, calculate the mass of water: 6.9 g.

Step3: Heat liquid water from $T_1$ to 100°C

$q_1 = m\times C_{liquid}\times\Delta T_1$
$\Delta T_1=100 - 82.56 = 17.44^\circ C$
$q_1=6.9\ g\times4.184\ J/g^\circ C\times17.44^\circ C$
$q_1=6.9\times4.184\times17.44\ J$
Calculate: 6.9*4.184 = 28.8696; 28.8696*17.44 ≈ 503.58 J ≈ 0.5036 kJ

Step4: Vaporize liquid water to steam (phase change)

Molar mass of water ($H_2O$) is 18.015 g/mol.
Moles of water: $n=\frac{6.9\ g}{18.015\ g/mol}\approx0.383\ mol$
$q_2 = n\times\Delta H_{vap}$
$q_2=0.383\ mol\times40.6\ kJ/mol\approx15.55\ kJ$

Step5: Heat steam from 100°C to $T_2$

$\Delta T_2=121.11 - 100 = 21.11^\circ C$
$q_3 = m\times C_{gas}\times\Delta T_2$
$q_3=6.9\ g\times2.03\ J/g^\circ C\times21.11^\circ C$
$q_3=6.9\times2.03\times21.11\ J$
Calculate: 6.9*2.03 = 14.007; 14.007*21.11 ≈ 295.69 J ≈ 0.2957 kJ

Step6: Total heat

$q_{total}=q_1 + q_2 + q_3$
$q_{total}=0.5036 + 15.55 + 0.2957\ kJ$
$q_{total}\approx16.3493\ kJ$ Wait, that can't be right. Wait, maybe I made a mistake in the phase. Wait, wait, 180.6°F is 82.56°C, 250°F is 121.11°C. Wait, but water at 82.56°C is liquid, and at 121.11°C is gas (steam), but to get from liquid at 82.56 to gas at 121.11, we need to:

1. Heat liquid from 82.56 to 100°C (liquid phase, $C_{liquid}$)
2. Vaporize liquid to gas at 100°C (phase change, $\Delta H_{vap}$)
3. Heat gas from 100°C to 121.11°C (gas phase, $C_{gas}$)

But wait, maybe the problem is that the temperature range is from 180.6°F (82.56°C) to 250.0°F (121.11°C), but maybe the water is already steam? No, 82.56°C is below 100°C, so it's liquid. Wait, but maybe I misread the heating curve. Wait, the heating curve has temperature on the x-axis. Wait, the first point is at -100°C (solid), then a phase change (melting) to liquid, then heating liquid to 100°C, then phase change (vaporization) to gas, then heating gas. So from 82.56°C (liquid) to 121.11°C (gas), so the steps are:

Liquid heating: 82.56 to 100°C

Vaporization: liquid to gas at 100°C

Gas heating: 100 to 121.11°C

But let's recalculate:

Step3: $q_1 = 6.9\ g \times 4.184\ J/g^\circ C \times (100 - 82.56)^\circ C$

100 - 82.56 = 17.44

6.9 * 4.184 = 28.8696

28.8696 * 17.44 = 28.8696 * 17 + 28.8696 * 0.44 = 490.7832 + 12.7026 = 503.4858 J ≈ 0.5035 kJ

Step4: Moles of water: 6.9 g / 18.015 g/mol ≈ 0.383 mol

$\Delta H_{vap} = 40.6\ kJ/mol$, so $q_2 = 0.383\ mol * 40.6\ kJ/mol ≈ 0.383 * 40.6 = 15.5498\ kJ ≈ 15.55\ kJ$

Step5: $q_3 = 6.9\ g * 2.03\ J/g^\circ C * (121.11 - 100)^\circ C$

121.11 - 100 = 21.11

6.9 * 2.03 = 14.007

14.007 * 21.11 = 14.007 * 20 + 14.007 * 1.11 = 280.14 + 15.5478 = 295.6878 J ≈ 0.2957 kJ

Total $q = 0.5035 + 15.55 + 0.2957 = 16.3492\ kJ ≈ 16.3\ kJ$? Wait, but that seems low. Wait, maybe I messed up the phase. Wait, maybe the initial temperature is 180.6°F, which is 82.56°C, and the final is 250°F, which is 121.11°C. But maybe the water is already in the gas phase? No, 82.56°C is below boiling, so it's liquid. Wait, maybe the heating curve is different. Wait, the table has $C_{solid}$, $C_{liquid}$, $C_{gas}$, $\Delta H_{fus}$, $\Delta H_{vap}$. So the process is:

Liquid (82.56°C) → Liquid (100°C): $q_1 = mC_{liquid}\Delta T$

Liquid (100°C) → Gas (100°C): $q_2 = n\Delta H_{vap}$

Gas (100°C) → Gas (121.11°C): $q_3 = mC_{gas}\Delta T$

Yes, that's correct.

Wait, but let's check the units. $C_{liquid}$ is 4.184 J/g°C, so q1 is in joules. Then q2 is in kJ (since $\Delta H_{vap}$ is kJ/mol), and q3 is in joules. So we need to convert q1 and q3 to kJ.

q1: 503.4858 J = 0.5034858 kJ

q3: 295.6878 J = 0.2956878 kJ

Then total q = 0.5034858 + 15.5498 + 0.2956878 ≈ 16.34897 kJ ≈ 16.3 kJ (rounded to one decimal place)

Wait, but maybe I made a mistake in the temperature conversion. Let's recheck the Fahrenheit to Celsius:

$^\circ C = (^\circ F - 32) \times \frac{5}{9}$

For 180.6°F: (180.6 - 32) = 148.6; 148.6 * 5/9 = 148.6 / 1.8 ≈ 82.555... ≈ 82.56°C (correct)

For 250.0°F: (250 - 32) = 218; 218 * 5/9 = 218 / 1.8 ≈ 121.111... ≈ 121.11°C (correct)

Another way: Maybe the problem is that the water is being heated from 180.6°F (which is 82.56°C) to 250.0°F (121.11°C), but maybe the phase is only gas? No, 82.56°C is liquid. Wait, maybe the heating curve is such that above 100°C is gas, so the steps are correct.

Alternatively, maybe the question is simpler: maybe the water is already in the gas phase? But 180.6°F is 82.56°C, which is below 100°C, so it's liquid. So the three steps are necessary.

Wait, but let's check the mass: 6.9 g. Let's recalculate q2:

Moles = 6.9 / 18.015 ≈ 0.383 mol

q2 = 0.383 * 40.6 ≈ 15.55 kJ (correct)

q1: 6.9 * 4.184 * (100 - 82.56) = 6.9 * 4.184 * 17.44 ≈ 6.9 * 72.96 ≈ 503.4 J ≈ 0.503 kJ (correct)

q3: 6.9 * 2.03 * (121.11 - 100) = 6.9 * 2.03 * 21.11 ≈ 6.9 * 42.85 ≈ 295.7 J ≈ 0.296 kJ (correct)

Total: 0.503 + 15.55 + 0.296 ≈ 16.349 kJ ≈ 16.3 kJ (one decimal place)

Wait, but maybe the problem is that the temperature range is from 180.6°F to 250.0°F, but the water is in the gas phase throughout? But 180.6°F is 82.56°C, which is below boiling, so it's liquid. So the phase change is necessary.

Alternatively, maybe I misread the heating curve. The heating curve has temperature on the x-axis, with points at -100°C (solid), then a phase change (melting) to liquid, then heating liquid to 100°C, then phase change (vaporization) to gas, then heating gas. So from 82.56°C (liquid) to 121.11°C (gas), the steps are liquid heating, vaporization, gas heating.

Yes, so the total energy is the sum of the three steps.

Answer:

16.3