Exampie 5.4 ron bars uy tha total mass of \( 3,5 \mathrm{~kg} \) are heated to 500 K and immediately placed in 28 kg water at 285 K Calculate the final temperature of the iron bars and the water. The heat capacity of steel is \( 46 \mathrm{~kJ} / \mathrm{kg} \mathrm{K} \).

To find the final temperature of the iron bars and the water, we can use the principle of conservation of energy, which states that the heat lost by the iron bars will be equal to the heat gained by the water. Let: - \( m_{\text{iron}} = 3.5 \, \text{kg} \) (mass of the iron bars) - \( C_{\text{iron}} = 46 \, \text{kJ/kg K} \) (heat capacity of the iron bars) - \( T_{\text{initial, iron}} = 500 \, \text{K} \) (initial temperature of the iron bars) - \( m_{\text{water}} = 28 \, \text{kg} \) (mass of the water) - \( C_{\text{water}} = 4.186 \, \text{kJ/kg K} \) (heat capacity of water) - \( T_{\text{initial, water}} = 285 \, \text{K} \) (initial temperature of the water) - \( T_f \) = final temperature of the system The heat lost by the iron bars can be expressed as: \[ Q_{\text{lost}} = m_{\text{iron}} \cdot C_{\text{iron}} \cdot (T_{\text{initial, iron}} - T_f) \] The heat gained by the water can be expressed as: \[ Q_{\text{gained}} = m_{\text{water}} \cdot C_{\text{water}} \cdot (T_f - T_{\text{initial, water}}) \] According to the conservation of energy: \[ Q_{\text{lost}} = Q_{\text{gained}} \] Substituting the expressions for \( Q_{\text{lost}} \) and \( Q_{\text{gained}} \): \[ m_{\text{iron}} \cdot C_{\text{iron}} \cdot (T_{\text{initial, iron}} - T_f) = m_{\text{water}} \cdot C_{\text{water}} \cdot (T_f - T_{\text{initial, water}}) \] Now substituting the known values: \[ 3.5 \, \text{kg} \cdot 46 \, \text{kJ/kg K} \cdot (500 \, \text{K} - T_f) = 28 \, \text{kg} \cdot 4.186 \, \text{kJ/kg K} \cdot (T_f - 285 \, \text{K}) \] Calculating the left side: \[ 3.5 \cdot 46 = 161 \, \text{kJ/K} \] So, \[ 161 \cdot (500 - T_f) = 28 \cdot 4.186 \cdot (T_f - 285) \] Calculating the right side: \[ 28 \cdot 4.186 = 117.208 \, \text{kJ/K} \] So, \[ 161 \cdot (500 - T_f) = 117.208 \cdot (T_f - 285) \] Expanding both sides: \[ 80500 - 161 T_f = 117.208 T_f - 33466.88 \] Rearranging the equation: \[ 80500 + 33466.88 = 161 T_f + 117.208 T_f \] \[ 113966.88 = 278.208 T_f \] Now, solving for \( T_f \): \[ T_f = \frac{113966.88}{278.208} \approx 409.5 \, \text{K} \] Thus, the final temperature of the iron bars and the water is approximately \( 409.5 \, \text{K} \).