/300 Resources Hint Submit A Question 3 of 3 $$ \stackrel{\infty}{.0} $$ At 1 atm , how much energy is required to heat $$ 75.0 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}(\mathrm{s}) $$ at $$ -10.0^{\circ} \mathrm{C} $$ to $$ \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) $$ at $$ 133.0^{\circ} \mathrm{C} $$ ? Use the heat transfer constants found in this table. $$ q= $$ $$ \square $$

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**Step 1. Heat the ice from $$-10.0^\circ\text{C}$$ to $$0.0^\circ\text{C}$$**

The energy required is given by
$$
q_1 = m \, c_{\text{ice}} \, \Delta T
$$
where
- $$m = 75.0\, \text{g}$$,
- $$c_{\text{ice}} = 2.09\, \text{J/(g}\cdot^\circ\text{C)}$$ (typical value),
- $$\Delta T = 0.0 - (-10.0) = 10.0^\circ\text{C}$$.

Thus,
$$
q_1 = 75.0\, \text{g} \times 2.09\, \frac{\text{J}}{\text{g}\cdot^\circ\text{C}} \times 10.0^\circ\text{C} = 1567.5\, \text{J}.
$$

---

**Step 2. Melt the ice at $$0.0^\circ\text{C}$$**

The heat required for melting is
$$
q_2 = n\, \Delta H_{\text{fus}},
$$
where
- The number of moles $$n$$ is calculated by
  $$
  n = \frac{75.0\, \text{g}}{18.0\, \text{g/mol}} \approx 4.17\, \text{mol}.
  $$
- The enthalpy of fusion is $$\Delta H_{\text{fus}} = 6.01\, \text{kJ/mol}$$.

Thus,
$$
q_2 = 4.17\, \text{mol} \times 6.01\, \text{kJ/mol} \approx 25.1\, \text{kJ} = 25100\, \text{J}.
$$

---

**Step 3. Heat the water from $$0.0^\circ\text{C}$$ to $$100.0^\circ\text{C}$$**

The energy required is
$$
q_3 = m \, c_{\text{water}} \, \Delta T,
$$
where
- $$c_{\text{water}} = 4.18\, \text{J/(g}\cdot^\circ\text{C)}$$,
- $$\Delta T = 100.0 - 0.0 = 100.0^\circ\text{C}$$.

Thus,
$$
q_3 = 75.0\, \text{g} \times 4.18\, \frac{\text{J}}{\text{g}\cdot^\circ\text{C}} \times 100.0^\circ\text{C} = 31350\, \text{J}.
$$

---

**Step 4. Vaporize the water at $$100.0^\circ\text{C}$$**

The energy required for vaporization is
$$
q_4 = n\, \Delta H_{\text{vap}},
$$
where $$\Delta H_{\text{vap}} = 40.7\, \text{kJ/mol}$$. Thus,
$$
q_4 = 4.17\, \text{mol} \times 40.7\, \text{kJ/mol} \approx 169.6\, \text{kJ} = 169600\, \text{J}.
$$

---

**Step 5. Heat the steam from $$100.0^\circ\text{C}$$ to $$133.0^\circ\text{C}$$**

The energy needed is
$$
q_5 = m \, c_{\text{steam}} \, \Delta T,
$$
where
- $$c_{\text{steam}} = 1.92\, \text{J/(g}\cdot^\circ\text{C)}$$,
- $$\Delta T = 133.0 - 100.0 = 33.0^\circ\text{C}$$.

Thus,
$$
q_5 = 75.0\, \text{g} \times 1.92\, \frac{\text{J}}{\text{g}\cdot^\circ\text{C}} \times 33.0^\circ\text{C} = 4752\, \text{J}.
$$

---

**Total Energy Required**

The total energy $$q_{\text{total}}$$ is the sum of all the steps:
$$
\begin{aligned}
q_{\text{total}} &= q_1 + q_2 + q_3 + q_4 + q_5 \
&= 1567.5\, \text{J} + 25100\, \text{J} + 31350\, \text{J} + 169600\, \text{J} + 4752\, \text{J} \
&\approx 232,369.5\, \text{J}.
\end{aligned}
$$

Thus, the energy required is approximately
$$
q \approx 2.32 \times 10^5\, \text{J}.
$$
The total energy required is approximately $$ 2.32 \times 10^5 $$ Joules.

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