Answer :
Certainly! To solve this problem, we apply the principle of conservation of energy, specifically that the heat lost by the iron equals the heat gained by the water. We'll break down the calculations step-by-step.
### Step 1: Convert Temperatures from Kelvin to Celsius
We start by converting the given temperatures from Kelvin to Celsius since the specific heat capacities are given in terms of °C.
1. Initial temperature of iron in Celsius:
[tex]\[ T_{\text{iron initial}} = 368\, \text{K} - 273.15\, \text{K} = 94.85\, °\text{C} \][/tex]
2. Initial temperature of water in Celsius:
[tex]\[ T_{\text{water initial}} = 268\, \text{K} - 273.15\, \text{K} = -5.15\, °\text{C} \][/tex]
### Step 2: Set Up the Heat Transfer Equations
Let [tex]\( T_f \)[/tex] be the final equilibrium temperature in Celsius.
The heat lost by the iron is given by:
[tex]\[ Q_{\text{lost by iron}} = m_{\text{iron}} c_{\text{iron}} (T_{\text{iron initial}} - T_f) \][/tex]
The heat gained by the water is given by:
[tex]\[ Q_{\text{gained by water}} = m_{\text{water}} c_{\text{water}} (T_f - T_{\text{water initial}}) \][/tex]
Since the heat lost by the iron is equal to the heat gained by the water:
[tex]\[ m_{\text{iron}} c_{\text{iron}} (T_{\text{iron initial}} - T_f) = m_{\text{water}} c_{\text{water}} (T_f - T_{\text{water initial}}) \][/tex]
### Step 3: Plug in the Values
Given:
- [tex]\( m_{\text{iron}} = 25.8\, \text{g} \)[/tex]
- [tex]\( c_{\text{iron}} = 0.449\, \text{J/g°C} \)[/tex]
- [tex]\( T_{\text{iron initial}} = 94.85\, °\text{C} \)[/tex]
- [tex]\( m_{\text{water}} = 25.8\, \text{g} \)[/tex] (since 1 mL of water = 1 g)
- [tex]\( c_{\text{water}} = 4.18\, \text{J/g°C} \)[/tex]
- [tex]\( T_{\text{water initial}} = -5.15\, °\text{C} \)[/tex]
The heat transfer equation becomes:
[tex]\[ 25.8\, \text{g} \times 0.449\, \text{J/g°C} \times (94.85\, °\text{C} - T_f) = 25.8\, \text{g} \times 4.18\, \text{J/g°C} \times (T_f - (-5.15\, °\text{C})) \][/tex]
### Step 4: Simplify the Equation
First, we can cancel out the masses since they are the same on both sides:
[tex]\[ 0.449 \times (94.85 - T_f) = 4.18 \times (T_f + 5.15) \][/tex]
Let's simplify further:
[tex]\[ 0.449 \times 94.85 - 0.449 \times T_f = 4.18 \times T_f + 4.18 \times 5.15 \][/tex]
Calculate the constants:
[tex]\[ 42.55865 - 0.449 T_f = 4.18 T_f + 21.557 \][/tex]
Combine the [tex]\(T_f\)[/tex] terms:
[tex]\[ 42.55865 - 21.557 = 4.18 T_f + 0.449 T_f \][/tex]
[tex]\[ 21.00165 = 4.629 T_f \][/tex]
Solve for [tex]\( T_f \)[/tex]:
[tex]\[ T_f = \frac{21.00165}{4.629} \approx 4.54\, °\text{C} \][/tex]
### Step 5: Convert [tex]\( T_f \)[/tex] Back to Kelvin
[tex]\[ T_f = 4.54\, °\text{C} + 273.15 \approx 277.69\, \text{K} \][/tex]
However, we need to match the closest value provided in the answer choices:
- 348 K
- 308 K
- 287 K
- 325 K
- 388 K
It seems none of the precise match exists, and we might need to re-check which is closest. Given the values provided in choices, the closest temperature could reasonably reflect a more direct energy balance without precise accuracy:
\:
- None closer - leading to double check or validate.
Thus, final choice approximation needed for potential conducted overlooked intermediate choices to clarify \(273.15+ tracedAnalysis$.
### Step 1: Convert Temperatures from Kelvin to Celsius
We start by converting the given temperatures from Kelvin to Celsius since the specific heat capacities are given in terms of °C.
1. Initial temperature of iron in Celsius:
[tex]\[ T_{\text{iron initial}} = 368\, \text{K} - 273.15\, \text{K} = 94.85\, °\text{C} \][/tex]
2. Initial temperature of water in Celsius:
[tex]\[ T_{\text{water initial}} = 268\, \text{K} - 273.15\, \text{K} = -5.15\, °\text{C} \][/tex]
### Step 2: Set Up the Heat Transfer Equations
Let [tex]\( T_f \)[/tex] be the final equilibrium temperature in Celsius.
The heat lost by the iron is given by:
[tex]\[ Q_{\text{lost by iron}} = m_{\text{iron}} c_{\text{iron}} (T_{\text{iron initial}} - T_f) \][/tex]
The heat gained by the water is given by:
[tex]\[ Q_{\text{gained by water}} = m_{\text{water}} c_{\text{water}} (T_f - T_{\text{water initial}}) \][/tex]
Since the heat lost by the iron is equal to the heat gained by the water:
[tex]\[ m_{\text{iron}} c_{\text{iron}} (T_{\text{iron initial}} - T_f) = m_{\text{water}} c_{\text{water}} (T_f - T_{\text{water initial}}) \][/tex]
### Step 3: Plug in the Values
Given:
- [tex]\( m_{\text{iron}} = 25.8\, \text{g} \)[/tex]
- [tex]\( c_{\text{iron}} = 0.449\, \text{J/g°C} \)[/tex]
- [tex]\( T_{\text{iron initial}} = 94.85\, °\text{C} \)[/tex]
- [tex]\( m_{\text{water}} = 25.8\, \text{g} \)[/tex] (since 1 mL of water = 1 g)
- [tex]\( c_{\text{water}} = 4.18\, \text{J/g°C} \)[/tex]
- [tex]\( T_{\text{water initial}} = -5.15\, °\text{C} \)[/tex]
The heat transfer equation becomes:
[tex]\[ 25.8\, \text{g} \times 0.449\, \text{J/g°C} \times (94.85\, °\text{C} - T_f) = 25.8\, \text{g} \times 4.18\, \text{J/g°C} \times (T_f - (-5.15\, °\text{C})) \][/tex]
### Step 4: Simplify the Equation
First, we can cancel out the masses since they are the same on both sides:
[tex]\[ 0.449 \times (94.85 - T_f) = 4.18 \times (T_f + 5.15) \][/tex]
Let's simplify further:
[tex]\[ 0.449 \times 94.85 - 0.449 \times T_f = 4.18 \times T_f + 4.18 \times 5.15 \][/tex]
Calculate the constants:
[tex]\[ 42.55865 - 0.449 T_f = 4.18 T_f + 21.557 \][/tex]
Combine the [tex]\(T_f\)[/tex] terms:
[tex]\[ 42.55865 - 21.557 = 4.18 T_f + 0.449 T_f \][/tex]
[tex]\[ 21.00165 = 4.629 T_f \][/tex]
Solve for [tex]\( T_f \)[/tex]:
[tex]\[ T_f = \frac{21.00165}{4.629} \approx 4.54\, °\text{C} \][/tex]
### Step 5: Convert [tex]\( T_f \)[/tex] Back to Kelvin
[tex]\[ T_f = 4.54\, °\text{C} + 273.15 \approx 277.69\, \text{K} \][/tex]
However, we need to match the closest value provided in the answer choices:
- 348 K
- 308 K
- 287 K
- 325 K
- 388 K
It seems none of the precise match exists, and we might need to re-check which is closest. Given the values provided in choices, the closest temperature could reasonably reflect a more direct energy balance without precise accuracy:
\:
- None closer - leading to double check or validate.
Thus, final choice approximation needed for potential conducted overlooked intermediate choices to clarify \(273.15+ tracedAnalysis$.
