91Ó°ÊÓ

The heat energy lost by the hot aluminum and iron pellets is equal to the heat gained by the water. We can write the energy conservation equation as follows: $$Q_{lost} = Q_{gained}$$

To calculate the heat energy lost by the aluminum pellets, we'll use the formula: $$Q_{aluminum} = m_{aluminum} \times c_{aluminum} \times (\Delta T)_{aluminum}$$ where \(Q_{aluminum}\) is the heat energy lost by the aluminum pellets, \(m_{aluminum}\) is the mass of the aluminum pellets, \(c_{aluminum}\) is the specific heat capacity of aluminum, and \((\Delta T)_{aluminum}\) is the change in temperature of the aluminum pellets. We have: \(m_{aluminum} = 5.00 \,\text{g}\) \(c_{aluminum} = 0.89 \,\mathrm{J} /{ }(^{\circ} \mathrm{C}) \cdot \mathrm{g}\) Initial temperature of the aluminum pellets, \(T_{initial}^{aluminum} = 100.0^{\circ} \mathrm{C}\) Final temperature of the aluminum pellets, \(T_{final}^{aluminum} = T_{final}\) (which we need to find) So, the temperature change is given by: \((\Delta T)_{aluminum} = T_{final} - T_{initial}^{aluminum}\)

To calculate the heat energy lost by the iron pellets, we'll use the formula: $$Q_{iron} = m_{iron} \times c_{iron} \times (\Delta T)_{iron}$$ where \(Q_{iron}\) is the heat energy lost by the iron pellets, \(m_{iron}\) is the mass of the iron pellets, \(c_{iron}\) is the specific heat capacity of iron, and \((\Delta T)_{iron}\) is the change in temperature of the iron pellets. We have: \(m_{iron} = 10.00 \,\mathrm{g}\) \(c_{iron} = 0.45 \,\mathrm{J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) Initial temperature of the iron pellets, \(T_{initial}^{iron} = 100.0^{\circ} \mathrm{C}\) Final temperature of the iron pellets, \(T_{final}^{iron} = T_{final}\) (which we need to find) So, the temperature change is given by: \((\Delta T)_{iron} = T_{final} - T_{initial}^{iron}\)

To calculate the heat energy gained by water, we'll use the formula: $$Q_{water} = m_{water} \times c_{water} \times (\Delta T)_{water}$$ where \(Q_{water}\) is the heat energy gained by water, \(m_{water}\) is the mass of the water, \(c_{water}\) is the specific heat capacity of water, and \((\Delta T)_{water}\) is the change in temperature of water. We have: \(m_{water} = 97.3 \,\mathrm{g}\) \(c_{water} = 4.18 \,\mathrm{J} / { }^{\circ} \mathrm{C} \cdot \mathrm{g}\) (approx) Initial temperature of water, \(T_{initial}^{water} = 22.0^{\circ} \mathrm{C}\) Final temperature of water, \(T_{final}^{water} = T_{final}\) (which we need to find) So, the temperature change is given by: \((\Delta T)_{water} = T_{final} - T_{initial}^{water}\)

Putting the expressions from steps 2, 3, and 4, we rewrite the energy conservation equation: $$m_{aluminum} \times c_{aluminum} \times (T_{final} - T_{initial}^{aluminum}) + m_{iron} \times c_{iron} \times (T_{final} - T_{initial}^{iron}) = m_{water} \times c_{water} \times (T_{final} - T_{initial}^{water})$$

Now we need to solve the equation in step 5 for \(T_{final}\): $$5.00 \times 0.89 \times (T_{final} - 100.0) + 10.00 \times 0.45 \times (T_{final} - 100.0) = 97.3 \times 4.18 \times (T_{final} - 22.0)$$ On solving for \(T_{final}\), we get: $$T_{final} \approx 26.0^{\circ} \mathrm{C}$$ Therefore, the final temperature of the metal and water mixture is approximately \(26.0^{\circ} \mathrm{C}\).