Answer :
Let’s solve this problem step-by-step by using the principles of thermodynamics and heat transfer.
### Step 1: Understand the given data
We are given:
- Mass of water ([tex]\( m_{\text{water}} \)[/tex]) = 500 grams
- Initial temperature of water ([tex]\( T_{\text{water initial}} \)[/tex]) = 56°C
- Specific heat capacity of water ([tex]\( c_{\text{water}} \)[/tex]) = 4.18 J/(g°C)
- Mass of ice ([tex]\( m_{\text{ice}} \)[/tex]) = 125 grams
- Initial temperature of ice ([tex]\( T_{\text{ice initial}} \)[/tex]) = 0°C
- Specific heat capacity of ice ([tex]\( c_{\text{ice}} \)[/tex]) = 2.09 J/(g°C)
- Heat of fusion for ice ([tex]\( H_{\text{fus}} \)[/tex]) = 334 J/g
### Step 2: Calculate the heat required to melt the ice
The heat required to melt the ice ([tex]\( q_{\text{melt ice}} \)[/tex]) is given by:
[tex]\[ q_{\text{melt ice}} = m_{\text{ice}} \times H_{\text{fus}} \][/tex]
Substituting the values:
[tex]\[ q_{\text{melt ice}} = 125 \, \text{grams} \times 334 \, \text{J/gram} = 41750 \, \text{J} \][/tex]
### Step 3: Set up the heat balance equation
Let the final temperature of the system be [tex]\( T_f \)[/tex].
The heat released by water as it cools down to the final temperature:
[tex]\[ q_{\text{water}} = m_{\text{water}} \times c_{\text{water}} \times (T_{\text{water initial}} - T_f) \][/tex]
The heat absorbed by the ice as it melts and warms up to the final temperature:
[tex]\[ q_{\text{ice}} = q_{\text{melt ice}} + m_{\text{ice}} \times c_{\text{water}} \times (T_f - 0) \][/tex]
At equilibrium, the heat lost by the water will be equal to the heat gained by the ice:
[tex]\[ m_{\text{water}} \times c_{\text{water}} \times (T_{\text{water initial}} - T_f) = q_{\text{melt ice}} + m_{\text{ice}} \times c_{\text{water}} \times T_f \][/tex]
### Step 4: Solve for the final temperature [tex]\( T_f \)[/tex]
Rearranging the above equation to solve for [tex]\( T_f \)[/tex]:
[tex]\[ m_{\text{water}} \times c_{\text{water}} \times T_{\text{water initial}} - m_{\text{water}} \times c_{\text{water}} \times T_f = q_{\text{melt ice}} + m_{\text{ice}} \times c_{\text{water}} \times T_f \][/tex]
Combine like terms:
[tex]\[ m_{\text{water}} \times c_{\text{water}} \times T_{\text{water initial}} = q_{\text{melt ice}} + (m_{\text{ice}} \times c_{\text{water}} + m_{\text{water}} \times c_{\text{water}}) \times T_f \][/tex]
Isolate [tex]\( T_f \)[/tex]:
[tex]\[ T_f = \frac{m_{\text{water}} \times c_{\text{water}} \times T_{\text{water initial}} - q_{\text{melt ice}}}{m_{\text{water}} \times c_{\text{water}} + m_{\text{ice}} \times c_{\text{water}}} \][/tex]
Substituting the known values:
[tex]\[ T_f = \frac{500 \, \text{grams} \times 4.18 \, \text{J/(g°C)} \times 56 \, \text{°C} - 41750 \, \text{J}}{(500 \, \text{grams} + 125 \, \text{grams}) \times 4.18 \, \text{J/(g°C)}} \][/tex]
Perform the calculations:
[tex]\[ T_f = \frac{500 \times 4.18 \times 56 - 41750}{625 \times 4.18} \approx 28.82 \, \text{°C} \][/tex]
### Final Answer:
The final temperature of the system, after the ice melts and reaches thermal equilibrium with the water, is approximately 28.82°C.
### Step 1: Understand the given data
We are given:
- Mass of water ([tex]\( m_{\text{water}} \)[/tex]) = 500 grams
- Initial temperature of water ([tex]\( T_{\text{water initial}} \)[/tex]) = 56°C
- Specific heat capacity of water ([tex]\( c_{\text{water}} \)[/tex]) = 4.18 J/(g°C)
- Mass of ice ([tex]\( m_{\text{ice}} \)[/tex]) = 125 grams
- Initial temperature of ice ([tex]\( T_{\text{ice initial}} \)[/tex]) = 0°C
- Specific heat capacity of ice ([tex]\( c_{\text{ice}} \)[/tex]) = 2.09 J/(g°C)
- Heat of fusion for ice ([tex]\( H_{\text{fus}} \)[/tex]) = 334 J/g
### Step 2: Calculate the heat required to melt the ice
The heat required to melt the ice ([tex]\( q_{\text{melt ice}} \)[/tex]) is given by:
[tex]\[ q_{\text{melt ice}} = m_{\text{ice}} \times H_{\text{fus}} \][/tex]
Substituting the values:
[tex]\[ q_{\text{melt ice}} = 125 \, \text{grams} \times 334 \, \text{J/gram} = 41750 \, \text{J} \][/tex]
### Step 3: Set up the heat balance equation
Let the final temperature of the system be [tex]\( T_f \)[/tex].
The heat released by water as it cools down to the final temperature:
[tex]\[ q_{\text{water}} = m_{\text{water}} \times c_{\text{water}} \times (T_{\text{water initial}} - T_f) \][/tex]
The heat absorbed by the ice as it melts and warms up to the final temperature:
[tex]\[ q_{\text{ice}} = q_{\text{melt ice}} + m_{\text{ice}} \times c_{\text{water}} \times (T_f - 0) \][/tex]
At equilibrium, the heat lost by the water will be equal to the heat gained by the ice:
[tex]\[ m_{\text{water}} \times c_{\text{water}} \times (T_{\text{water initial}} - T_f) = q_{\text{melt ice}} + m_{\text{ice}} \times c_{\text{water}} \times T_f \][/tex]
### Step 4: Solve for the final temperature [tex]\( T_f \)[/tex]
Rearranging the above equation to solve for [tex]\( T_f \)[/tex]:
[tex]\[ m_{\text{water}} \times c_{\text{water}} \times T_{\text{water initial}} - m_{\text{water}} \times c_{\text{water}} \times T_f = q_{\text{melt ice}} + m_{\text{ice}} \times c_{\text{water}} \times T_f \][/tex]
Combine like terms:
[tex]\[ m_{\text{water}} \times c_{\text{water}} \times T_{\text{water initial}} = q_{\text{melt ice}} + (m_{\text{ice}} \times c_{\text{water}} + m_{\text{water}} \times c_{\text{water}}) \times T_f \][/tex]
Isolate [tex]\( T_f \)[/tex]:
[tex]\[ T_f = \frac{m_{\text{water}} \times c_{\text{water}} \times T_{\text{water initial}} - q_{\text{melt ice}}}{m_{\text{water}} \times c_{\text{water}} + m_{\text{ice}} \times c_{\text{water}}} \][/tex]
Substituting the known values:
[tex]\[ T_f = \frac{500 \, \text{grams} \times 4.18 \, \text{J/(g°C)} \times 56 \, \text{°C} - 41750 \, \text{J}}{(500 \, \text{grams} + 125 \, \text{grams}) \times 4.18 \, \text{J/(g°C)}} \][/tex]
Perform the calculations:
[tex]\[ T_f = \frac{500 \times 4.18 \times 56 - 41750}{625 \times 4.18} \approx 28.82 \, \text{°C} \][/tex]
### Final Answer:
The final temperature of the system, after the ice melts and reaches thermal equilibrium with the water, is approximately 28.82°C.
