To find the final temperature when the iron bar is dropped into the water, we can use the principle of conservation of energy, which states that the heat lost by the iron will be equal to the heat gained by the water.
The specific heat capacities are as follows:
- Specific heat capacity of iron (\(c_{\text{iron}}\)) is approximately \(0.449 , \text{J/g°C}\)
- Specific heat capacity of water (\(c_{\text{water}}\)) is \(4.186 , \text{J/g°C}\)
First, we convert the masses of the materials to grams:
- Mass of iron (\(m_{\text{iron}}\)) = \(2.5 , \text{kg} = 2500 , \text{g}\)
- Mass of water (\(m_{\text{water}}\)) = \(0.24 , \text{kg} = 240 , \text{g}\)
Let \(T_f\) be the final temperature of both the iron and the water.
The heat lost by the iron bar is given by: \[ Q_{\text{lost}} = m_{\text{iron}} \cdot c_{\text{iron}} \cdot (T_{\text{initial,iron}} - T_f) = 2500 , \text{g} \cdot 0.449 , \text{J/g°C} \cdot (94°C - T_f) \]
The heat gained by the water is given by: \[ Q_{\text{gained}} = m_{\text{water}} \cdot c_{\text{water}} \cdot (T_f - T_{\text{initial,water}}) = 240 , \text{g} \cdot 4.186 , \text{J/g°C} \cdot (T_f - 20°C) \]
Setting \(Q_{\text{lost}} = Q_{\text{gained}}\): \[ 2500 \cdot 0.449 \cdot (94 - T_f) = 240 \cdot 4.186 \cdot (T_f - 20) \]
Calculating the constants: \[ 2500 \cdot 0.449 = 1122.5 \] \[ 240 \cdot 4.186 = 1004.64 \]
Now substituting: \[ 1122.5 \cdot (94 - T_f) = 1004.64 \cdot (T_f - 20) \]
Expanding both sides: \[ 105605 - 1122.5 T_f = 1004.64 T_f - 20092.8 \]
Combining like terms: \[ 105605 + 20092.8 = 1004.64 T_f + 1122.5 T_f \] \[ 125697.8 = 2127.14 T_f \]
Now solving for \(T_f\): \[ T_f = \frac{125697.8}{2127.14} \approx 59.05°C \]
Thus, the final temperature when the iron bar is dropped into the water is approximately 59.1°C.
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