Question

a 0.588 g sample of steam at 103.2 °c is condensed into a container with 4.91 g of water at 15.8 °c. what is the final temperature of the water mixture if no heat is lost? the specific heat of water is 4.18 \\(\frac{j}{g\cdot ^\circ c}\\), the specific heat of steam is 2.01 \\(\frac{j}{g\cdot ^\circ c}\\), and \\(\delta h_{vap} = 40.7\\) kj/mol. \\(t_f = \\) \\(^\circ c\\)

Explanation:

Step1: Convert steam mass to moles

Molar mass of $\text{H}_2\text{O} = 18.016\ \text{g/mol}$
$\text{Moles of steam} = \frac{0.588\ \text{g}}{18.016\ \text{g/mol}} \approx 0.03264\ \text{mol}$

Step2: Define heat transfer balance

Heat lost by steam = Heat gained by cold water
$q_{\text{steam cool}} + q_{\text{condensation}} + q_{\text{hot water cool}} = q_{\text{cold water warm}}$

Step3: Write heat equations

Let $T_f$ = final temperature

1. Cool steam to 100°C:

$q_1 = m_{\text{steam}} \times c_{\text{steam}} \times (103.2^\circ\text{C} - 100^\circ\text{C}) = 0.588\ \text{g} \times 2.01\ \frac{\text{J}}{\text{g·°C}} \times 3.2^\circ\text{C} \approx 3.77\ \text{J}$

1. Condense steam:

$q_2 = n \times \Delta H_{\text{vap}} = 0.03264\ \text{mol} \times 40700\ \frac{\text{J}}{\text{mol}} \approx 1328.4\ \text{J}$

1. Cool condensed steam to $T_f$:

$q_3 = m_{\text{steam}} \times c_{\text{water}} \times (100^\circ\text{C} - T_f) = 0.588\ \text{g} \times 4.18\ \frac{\text{J}}{\text{g·°C}} \times (100 - T_f) = 2.458(100 - T_f)\ \text{J}$

1. Warm cold water to $T_f$:

$q_4 = m_{\text{cold water}} \times c_{\text{water}} \times (T_f - 15.8^\circ\text{C}) = 4.91\ \text{g} \times 4.18\ \frac{\text{J}}{\text{g·°C}} \times (T_f - 15.8) = 20.52(T_f - 15.8)\ \text{J}$

Step4: Combine and solve for $T_f$

$3.77 + 1328.4 + 2.458(100 - T_f) = 20.52(T_f - 15.8)$
$1332.17 + 245.8 - 2.458T_f = 20.52T_f - 324.22$
$1577.97 + 324.22 = 20.52T_f + 2.458T_f$
$1902.19 = 22.978T_f$
$T_f = \frac{1902.19}{22.978} \approx 82.8^\circ\text{C}$

Answer:

$82.8$ °C