How much heat is released when 105 g of steam at 100.0 C is cooled to ice at -15.0 C? The enthalpy of vaporization of water is 40.67 kj/mol, the enthalpy of fusion for water is 6.01 kj/mol, the specific heat capacity of liquid water is 4.18 J/g C and the specific heat capacity of ice is 2.02 J/g C. (q heating = m x C x ΔT, q phase change = ΔH vap x n)
How much heat is released when 105 g of steam at 100.0 C is cooled to ice at -15.0 C? The enthalpy of vaporization of water is 40.67 kj/mol, the enthalpy of fusion for water is 6.01 kj/mol, the specific heat capacity of liquid water is 4.18 J/g C and the specific heat capacity of ice is 2.02 J/g C. (q heating = m x C x ΔT, q phase change = ΔH vap x n)
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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How much heat is released when 105 g of steam at 100.0 C is cooled to ice at -15.0 C? The enthalpy
of vaporization of water is 40.67 kj/mol, the enthalpy of fusion for water is 6.01 kj/mol, the specific heat
capacity of liquid water is 4.18 J/g C and the specific heat capacity of ice is 2.02 J/g C.
(q heating = m x C x ΔT, q phase change = ΔH vap x n)
of vaporization of water is 40.67 kj/mol, the enthalpy of fusion for water is 6.01 kj/mol, the specific heat
capacity of liquid water is 4.18 J/g C and the specific heat capacity of ice is 2.02 J/g C.
(q heating = m x C x ΔT, q phase change = ΔH vap x n)
![**Problem 7: How Much Heat is Released?**
**Question:**
How much heat is released when 105 g of steam at 100.0°C is cooled to ice at -15.0°C?
**Given data:**
- The enthalpy of vaporization of water (ΔHvap) is 40.67 kJ/mol.
- The enthalpy of fusion for water (ΔHfus) is 6.01 kJ/mol.
- The specific heat capacity (C) of liquid water is 4.18 J/g°C.
- The specific heat capacity (C) of ice is 2.02 J/g°C.
**Formulas to use:**
- For heating/cooling (q_heating):
\[ q_{\text{heating}} = m \times C \times \Delta T \]
- For phase changes (q_phase change):
\[ q_{\text{phase change}} = \Delta H_{\text{vap}} \times n \]
**Note:**
- \( m \) is the mass of the substance.
- \( C \) is the specific heat capacity.
- \( \Delta T \) is the change in temperature.
- \( \Delta H_{\text{vap}} \) is the enthalpy of vaporization.
- \( n \) is the number of moles.
**Explanation:**
To solve this problem, we must calculate the amount of heat released in each step of the cooling process, which involves multiple phase changes and temperature adjustments from steam to liquid water, then from liquid water to ice. Each step involves cooling within a phase or transitioning between phases (vaporization or fusion).
1. **Condensation:** Cooling the steam to liquid water at 100.0°C using the enthalpy of vaporization.
2. **Cooling Liquid Water:** Further cooling liquid water from 100.0°C to 0.0°C.
3. **Freezing:** Transition from liquid water at 0.0°C to ice using the enthalpy of fusion.
4. **Cooling Ice:** Cooling the ice from 0.0°C to -15.0°C using the specific heat capacity of ice.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F85650e2f-17ad-4aa4-9d56-acd675fbc675%2Fd461c043-3717-459e-b692-adc5e124f87e%2Fk3bvw96_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem 7: How Much Heat is Released?**
**Question:**
How much heat is released when 105 g of steam at 100.0°C is cooled to ice at -15.0°C?
**Given data:**
- The enthalpy of vaporization of water (ΔHvap) is 40.67 kJ/mol.
- The enthalpy of fusion for water (ΔHfus) is 6.01 kJ/mol.
- The specific heat capacity (C) of liquid water is 4.18 J/g°C.
- The specific heat capacity (C) of ice is 2.02 J/g°C.
**Formulas to use:**
- For heating/cooling (q_heating):
\[ q_{\text{heating}} = m \times C \times \Delta T \]
- For phase changes (q_phase change):
\[ q_{\text{phase change}} = \Delta H_{\text{vap}} \times n \]
**Note:**
- \( m \) is the mass of the substance.
- \( C \) is the specific heat capacity.
- \( \Delta T \) is the change in temperature.
- \( \Delta H_{\text{vap}} \) is the enthalpy of vaporization.
- \( n \) is the number of moles.
**Explanation:**
To solve this problem, we must calculate the amount of heat released in each step of the cooling process, which involves multiple phase changes and temperature adjustments from steam to liquid water, then from liquid water to ice. Each step involves cooling within a phase or transitioning between phases (vaporization or fusion).
1. **Condensation:** Cooling the steam to liquid water at 100.0°C using the enthalpy of vaporization.
2. **Cooling Liquid Water:** Further cooling liquid water from 100.0°C to 0.0°C.
3. **Freezing:** Transition from liquid water at 0.0°C to ice using the enthalpy of fusion.
4. **Cooling Ice:** Cooling the ice from 0.0°C to -15.0°C using the specific heat capacity of ice.
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Step 1: Heat required to convert 100°C steam to 100°C liquid water
VIEWStep 2: Heat required to convert 100°C liquid water to 0°C liquid water
VIEWStep 3: Heat required to convert 0°C liquid water to 0°C ice
VIEWStep 4: Heat required to convert 0°C ice to -15°C ice
VIEWStep 5: Total heat required
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