Answer :
To solve this problem, we need to account for the various heat transfers involved. We'll calculate the heat required to raise the temperature of both the aluminum calorimeter and the water, then determine how much steam is needed to provide this heat.
### Given Data:
- Specific heat capacity of water, [tex]\( c_{\text{water}} = 4.186 \ \text{J/g°C} \)[/tex]
- Specific heat capacity of aluminum, [tex]\( c_{\text{aluminum}} = 0.897 \ \text{J/g°C} \)[/tex]
- Specific heat capacity of steam, [tex]\( c_{\text{steam}} = 2.03 \ \text{J/g°C} \)[/tex]
- Latent heat of vaporization of water, [tex]\( L_{\text{vapor}} = 2260 \ \text{J/g} \)[/tex]
- Mass of aluminum, [tex]\( m_{\text{aluminum}} = 120 \ \text{g} \)[/tex]
- Mass of water, [tex]\( m_{\text{water}} = 390 \ \text{g} \)[/tex]
- Initial temperature, [tex]\( T_{\text{initial}} = 15 \ \text{°C} \)[/tex]
- Final temperature, [tex]\( T_{\text{final}} = 62 \ \text{°C} \)[/tex]
- Initial temperature of steam, [tex]\( T_{\text{steam initial}} = 110 \ \text{°C} \)[/tex]
- Condensation temperature of steam, [tex]\( T_{\text{steam condensed}} = 100 \ \text{°C} \)[/tex]
### Step-by-Step Solution:
1. Calculate Heat Required to Raise Temperature of Aluminum:
- [tex]\( Q_{\text{aluminum}} = m_{\text{aluminum}} \cdot c_{\text{aluminum}} \cdot (T_{\text{final}} - T_{\text{initial}}) \)[/tex]
- [tex]\( Q_{\text{aluminum}} = 120 \ \text{g} \cdot 0.897 \ \text{J/g°C} \cdot (62 - 15) \ \text{°C} \)[/tex]
- [tex]\( Q_{\text{aluminum}} = 120 \cdot 0.897 \cdot 47 \)[/tex]
- [tex]\( Q_{\text{aluminum}} = 5046.72 \ \text{J} \)[/tex]
2. Calculate Heat Required to Raise Temperature of Water:
- [tex]\( Q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot (T_{\text{final}} - T_{\text{initial}}) \)[/tex]
- [tex]\( Q_{\text{water}} = 390 \ \text{g} \cdot 4.186 \ \text{J/g°C} \cdot (62 - 15) \ \text{°C} \)[/tex]
- [tex]\( Q_{\text{water}} = 390 \cdot 4.186 \cdot 47 \)[/tex]
- [tex]\( Q_{\text{water}} = 76703.22 \ \text{J} \)[/tex]
3. Calculate Total Heat Required:
- [tex]\( Q_{\text{needed}} = Q_{\text{aluminum}} + Q_{\text{water}} \)[/tex]
- [tex]\( Q_{\text{needed}} = 5046.72 \ \text{J} + 76703.22 \ \text{J} \)[/tex]
- [tex]\( Q_{\text{needed}} = 81749.94 \ \text{J} \)[/tex]
4. Calculate Heat Released by Steam Cooling Down to 100°C:
- [tex]\( Q_{\text{steam cooling}} = m_{\text{steam}} \cdot c_{\text{steam}} \cdot (T_{\text{steam initial}} - T_{\text{steam condensed}}) \)[/tex]
- [tex]\( Q_{\text{steam cooling}} = m_{\text{steam}} \cdot 2.03 \ \text{J/g°C} \cdot (110 - 100) \ \text{°C} \)[/tex]
- [tex]\( Q_{\text{steam cooling}} = m_{\text{steam}} \cdot 20.3 \ \text{J/g} \)[/tex]
5. Calculate Heat Released by Steam Condensing:
- [tex]\( Q_{\text{steam condensing}} = m_{\text{steam}} \cdot L_{\text{vapor}} \)[/tex]
- [tex]\( Q_{\text{steam condensing}} = m_{\text{steam}} \cdot 2260 \ \text{J/g} \)[/tex]
6. Calculate Heat Released by Condensed Steam Cooling to 62°C:
- [tex]\( Q_{\text{condensed steam cooling}} = m_{\text{steam}} \cdot c_{\text{water}} \cdot (T_{\text{final}} - T_{\text{steam condensed}}) \)[/tex]
- [tex]\( Q_{\text{condensed steam cooling}} = m_{\text{steam}} \cdot 4.186 \ \text{J/g°C} \cdot (62 - 100) \ \text{°C} \)[/tex]
- [tex]\( Q_{\text{condensed steam cooling}} = m_{\text{steam}} \cdot 4.186 \cdot (-38) \ \text{°C} \)[/tex]
- [tex]\( Q_{\text{condensed steam cooling}} = m_{\text{steam}} \cdot (-159.068) \ \text{J/g} \)[/tex]
7. Total Heat Released by Each Gram of Steam:
- [tex]\( Q_{\text{per gram steam}} = Q_{\text{steam cooling}} + Q_{\text{steam condensing}} + Q_{\text{condensed steam cooling}} \)[/tex]
- [tex]\( Q_{\text{per gram steam}} = 20.3 \ \text{J/g} + 2260 \ \text{J/g} + (-159.068 \ \text{J/g}) \)[/tex]
- [tex]\( Q_{\text{per gram steam}} = 2121.232 \ \text{J/g} \)[/tex]
8. Calculate the Mass of Steam Needed:
- [tex]\( m_{\text{steam needed}} = \frac{Q_{\text{needed}}}{Q_{\text{per gram steam}}} \)[/tex]
- [tex]\( m_{\text{steam needed}} = \frac{81749.94 \ \text{J}}{2121.232 \ \text{J/g}} \)[/tex]
- [tex]\( m_{\text{steam needed}} \approx 38.55 \ \text{g} \)[/tex]
### Final Answer:
Approximately 38.55 grams of steam at 110°C are needed to raise the temperature of the water and the aluminum calorimeter from 15°C to 62°C.
### Given Data:
- Specific heat capacity of water, [tex]\( c_{\text{water}} = 4.186 \ \text{J/g°C} \)[/tex]
- Specific heat capacity of aluminum, [tex]\( c_{\text{aluminum}} = 0.897 \ \text{J/g°C} \)[/tex]
- Specific heat capacity of steam, [tex]\( c_{\text{steam}} = 2.03 \ \text{J/g°C} \)[/tex]
- Latent heat of vaporization of water, [tex]\( L_{\text{vapor}} = 2260 \ \text{J/g} \)[/tex]
- Mass of aluminum, [tex]\( m_{\text{aluminum}} = 120 \ \text{g} \)[/tex]
- Mass of water, [tex]\( m_{\text{water}} = 390 \ \text{g} \)[/tex]
- Initial temperature, [tex]\( T_{\text{initial}} = 15 \ \text{°C} \)[/tex]
- Final temperature, [tex]\( T_{\text{final}} = 62 \ \text{°C} \)[/tex]
- Initial temperature of steam, [tex]\( T_{\text{steam initial}} = 110 \ \text{°C} \)[/tex]
- Condensation temperature of steam, [tex]\( T_{\text{steam condensed}} = 100 \ \text{°C} \)[/tex]
### Step-by-Step Solution:
1. Calculate Heat Required to Raise Temperature of Aluminum:
- [tex]\( Q_{\text{aluminum}} = m_{\text{aluminum}} \cdot c_{\text{aluminum}} \cdot (T_{\text{final}} - T_{\text{initial}}) \)[/tex]
- [tex]\( Q_{\text{aluminum}} = 120 \ \text{g} \cdot 0.897 \ \text{J/g°C} \cdot (62 - 15) \ \text{°C} \)[/tex]
- [tex]\( Q_{\text{aluminum}} = 120 \cdot 0.897 \cdot 47 \)[/tex]
- [tex]\( Q_{\text{aluminum}} = 5046.72 \ \text{J} \)[/tex]
2. Calculate Heat Required to Raise Temperature of Water:
- [tex]\( Q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot (T_{\text{final}} - T_{\text{initial}}) \)[/tex]
- [tex]\( Q_{\text{water}} = 390 \ \text{g} \cdot 4.186 \ \text{J/g°C} \cdot (62 - 15) \ \text{°C} \)[/tex]
- [tex]\( Q_{\text{water}} = 390 \cdot 4.186 \cdot 47 \)[/tex]
- [tex]\( Q_{\text{water}} = 76703.22 \ \text{J} \)[/tex]
3. Calculate Total Heat Required:
- [tex]\( Q_{\text{needed}} = Q_{\text{aluminum}} + Q_{\text{water}} \)[/tex]
- [tex]\( Q_{\text{needed}} = 5046.72 \ \text{J} + 76703.22 \ \text{J} \)[/tex]
- [tex]\( Q_{\text{needed}} = 81749.94 \ \text{J} \)[/tex]
4. Calculate Heat Released by Steam Cooling Down to 100°C:
- [tex]\( Q_{\text{steam cooling}} = m_{\text{steam}} \cdot c_{\text{steam}} \cdot (T_{\text{steam initial}} - T_{\text{steam condensed}}) \)[/tex]
- [tex]\( Q_{\text{steam cooling}} = m_{\text{steam}} \cdot 2.03 \ \text{J/g°C} \cdot (110 - 100) \ \text{°C} \)[/tex]
- [tex]\( Q_{\text{steam cooling}} = m_{\text{steam}} \cdot 20.3 \ \text{J/g} \)[/tex]
5. Calculate Heat Released by Steam Condensing:
- [tex]\( Q_{\text{steam condensing}} = m_{\text{steam}} \cdot L_{\text{vapor}} \)[/tex]
- [tex]\( Q_{\text{steam condensing}} = m_{\text{steam}} \cdot 2260 \ \text{J/g} \)[/tex]
6. Calculate Heat Released by Condensed Steam Cooling to 62°C:
- [tex]\( Q_{\text{condensed steam cooling}} = m_{\text{steam}} \cdot c_{\text{water}} \cdot (T_{\text{final}} - T_{\text{steam condensed}}) \)[/tex]
- [tex]\( Q_{\text{condensed steam cooling}} = m_{\text{steam}} \cdot 4.186 \ \text{J/g°C} \cdot (62 - 100) \ \text{°C} \)[/tex]
- [tex]\( Q_{\text{condensed steam cooling}} = m_{\text{steam}} \cdot 4.186 \cdot (-38) \ \text{°C} \)[/tex]
- [tex]\( Q_{\text{condensed steam cooling}} = m_{\text{steam}} \cdot (-159.068) \ \text{J/g} \)[/tex]
7. Total Heat Released by Each Gram of Steam:
- [tex]\( Q_{\text{per gram steam}} = Q_{\text{steam cooling}} + Q_{\text{steam condensing}} + Q_{\text{condensed steam cooling}} \)[/tex]
- [tex]\( Q_{\text{per gram steam}} = 20.3 \ \text{J/g} + 2260 \ \text{J/g} + (-159.068 \ \text{J/g}) \)[/tex]
- [tex]\( Q_{\text{per gram steam}} = 2121.232 \ \text{J/g} \)[/tex]
8. Calculate the Mass of Steam Needed:
- [tex]\( m_{\text{steam needed}} = \frac{Q_{\text{needed}}}{Q_{\text{per gram steam}}} \)[/tex]
- [tex]\( m_{\text{steam needed}} = \frac{81749.94 \ \text{J}}{2121.232 \ \text{J/g}} \)[/tex]
- [tex]\( m_{\text{steam needed}} \approx 38.55 \ \text{g} \)[/tex]
### Final Answer:
Approximately 38.55 grams of steam at 110°C are needed to raise the temperature of the water and the aluminum calorimeter from 15°C to 62°C.
