Answer :
Sure! Let's walk through the essential steps to solve this problem. The final goal is to find the final equilibrium temperature ([tex]\( T_f \)[/tex]) of the water when the iron and water reach thermal equilibrium.
### Step-by-Step Solution:
1. Convert Initial Temperatures from Kelvin to Celsius:
- For iron:
[tex]\[ T_{initial, iron} = 368 \, \text{K} - 273.15 = 94.85 \, ^{\circ}C \][/tex]
- For water:
[tex]\[ T_{initial, water} = 268 \, \text{K} - 273.15 = -5.15 \, ^{\circ}C \][/tex]
2. Calculate the Mass of Water:
Given the volume of water ([tex]\( V_{water} \)[/tex]) is 25.8 mL and the density ([tex]\( \rho_{water} \)[/tex]) is approximately 1.0 g/mL, the mass of water ([tex]\( m_{water} \)[/tex]) is:
[tex]\[ m_{water} = V_{water} \times \rho_{water} = 25.8 \, \text{mL} \times 1.0 \, \text{g/mL} = 25.8 \, \text{g} \][/tex]
3. Set Up the Energy Balance Equation:
According to the conservation of energy, the heat lost by iron ([tex]\( Q_{iron} \)[/tex]) will be equal to the heat gained by water ([tex]\( Q_{water} \)[/tex]):
[tex]\[ Q_{iron} = Q_{water} \][/tex]
The formula for heat ([tex]\( Q \)[/tex]) is:
[tex]\[ Q = m \times c \times \Delta T \][/tex]
where [tex]\( m \)[/tex] is the mass, [tex]\( c \)[/tex] is the specific heat capacity, and [tex]\( \Delta T \)[/tex] is the change in temperature.
For iron (losing heat):
[tex]\[ Q_{iron} = m_{iron} \times c_{iron} \times (T_{initial, iron} - T_f) \][/tex]
For water (gaining heat):
[tex]\[ Q_{water} = m_{water} \times c_{water} \times (T_f - T_{initial, water}) \][/tex]
Equating the two expressions:
[tex]\[ m_{iron} \times c_{iron} \times (T_{initial, iron} - T_f) = m_{water} \times c_{water} \times (T_f - T_{initial, water}) \][/tex]
4. Rearrange the Equation to Solve for [tex]\( T_f \)[/tex]:
Plugging in the given values:
[tex]\[ 25.0 \, \text{g} \times 0.449 \, \text{J/g}^{\circ}\text{C} \times (94.85 \, ^{\circ}\text{C} - T_f) = 25.8 \, \text{g} \times 4.18 \, \text{J/g}^{\circ}\text{C} \times (T_f - (-5.15 \, ^{\circ}\text{C})) \][/tex]
Simplify the equation:
[tex]\[ 25.0 \times 0.449 \times (94.85 - T_f) = 25.8 \times 4.18 \times (T_f + 5.15) \][/tex]
5. Calculate the Final Temperature [tex]\( T_f \)[/tex]:
Solving this equation numerically, we find:
[tex]\[ T_f \approx 4.28 \, ^{\circ}\text{C} \][/tex]
Convert this back to Kelvin:
[tex]\[ T_f = 4.28 \, ^{\circ}\text{C} + 273.15 = 277.43 \, \text{K} \][/tex]
### Final Answer:
So, the final temperature of the water and the iron when they reach thermal equilibrium will be approximately:
[tex]\[ 277 \, \text{K} \approx 277 \, \text{K} \][/tex]
Hence, the correct option is not precisely given in your options, but the closest value to the calculated result would be around [tex]\( 277 \, \text{K} \)[/tex].
### Step-by-Step Solution:
1. Convert Initial Temperatures from Kelvin to Celsius:
- For iron:
[tex]\[ T_{initial, iron} = 368 \, \text{K} - 273.15 = 94.85 \, ^{\circ}C \][/tex]
- For water:
[tex]\[ T_{initial, water} = 268 \, \text{K} - 273.15 = -5.15 \, ^{\circ}C \][/tex]
2. Calculate the Mass of Water:
Given the volume of water ([tex]\( V_{water} \)[/tex]) is 25.8 mL and the density ([tex]\( \rho_{water} \)[/tex]) is approximately 1.0 g/mL, the mass of water ([tex]\( m_{water} \)[/tex]) is:
[tex]\[ m_{water} = V_{water} \times \rho_{water} = 25.8 \, \text{mL} \times 1.0 \, \text{g/mL} = 25.8 \, \text{g} \][/tex]
3. Set Up the Energy Balance Equation:
According to the conservation of energy, the heat lost by iron ([tex]\( Q_{iron} \)[/tex]) will be equal to the heat gained by water ([tex]\( Q_{water} \)[/tex]):
[tex]\[ Q_{iron} = Q_{water} \][/tex]
The formula for heat ([tex]\( Q \)[/tex]) is:
[tex]\[ Q = m \times c \times \Delta T \][/tex]
where [tex]\( m \)[/tex] is the mass, [tex]\( c \)[/tex] is the specific heat capacity, and [tex]\( \Delta T \)[/tex] is the change in temperature.
For iron (losing heat):
[tex]\[ Q_{iron} = m_{iron} \times c_{iron} \times (T_{initial, iron} - T_f) \][/tex]
For water (gaining heat):
[tex]\[ Q_{water} = m_{water} \times c_{water} \times (T_f - T_{initial, water}) \][/tex]
Equating the two expressions:
[tex]\[ m_{iron} \times c_{iron} \times (T_{initial, iron} - T_f) = m_{water} \times c_{water} \times (T_f - T_{initial, water}) \][/tex]
4. Rearrange the Equation to Solve for [tex]\( T_f \)[/tex]:
Plugging in the given values:
[tex]\[ 25.0 \, \text{g} \times 0.449 \, \text{J/g}^{\circ}\text{C} \times (94.85 \, ^{\circ}\text{C} - T_f) = 25.8 \, \text{g} \times 4.18 \, \text{J/g}^{\circ}\text{C} \times (T_f - (-5.15 \, ^{\circ}\text{C})) \][/tex]
Simplify the equation:
[tex]\[ 25.0 \times 0.449 \times (94.85 - T_f) = 25.8 \times 4.18 \times (T_f + 5.15) \][/tex]
5. Calculate the Final Temperature [tex]\( T_f \)[/tex]:
Solving this equation numerically, we find:
[tex]\[ T_f \approx 4.28 \, ^{\circ}\text{C} \][/tex]
Convert this back to Kelvin:
[tex]\[ T_f = 4.28 \, ^{\circ}\text{C} + 273.15 = 277.43 \, \text{K} \][/tex]
### Final Answer:
So, the final temperature of the water and the iron when they reach thermal equilibrium will be approximately:
[tex]\[ 277 \, \text{K} \approx 277 \, \text{K} \][/tex]
Hence, the correct option is not precisely given in your options, but the closest value to the calculated result would be around [tex]\( 277 \, \text{K} \)[/tex].
