Question

A \(5.00-\mathrm{g}\) sample of aluminum pellets (specific heat capacity \(=\) \(0.89 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) ) and a \(10.00-\mathrm{g}\) sample of iron pellets (specific heat capacity \(=0.45 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) ) are heated to \(100.0^{\circ} \mathrm{C}\). The mixture of hot iron and aluminum is then dropped into \(97.3 \mathrm{~g}\) water at \(22.0^{\circ} \mathrm{C}\). Calculate the final temperature of the metal and water mixture, assuming no heat loss to the surroundings.

Short answer

The final temperature of the metal and water mixture is approximately \(26.0^{\circ} \mathrm{C}\).

Step-by-step solution

1. Write down the energy conservation equation

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: $$Q_{lost} = Q_{gained}$$

2. Calculate the heat energy lost by aluminum pellets

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: \((\Delta T)_{aluminum} = T_{final} - T_{initial}^{aluminum}\)

3. Calculate the heat energy lost by iron pellets

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: \((\Delta T)_{iron} = T_{final} - T_{initial}^{iron}\)

4. Calculate the heat energy gained by water

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: \((\Delta T)_{water} = T_{final} - T_{initial}^{water}\)

5. Write down the energy conservation equation with found expressions

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

6. Solve for the final temperature

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 \(26.0^{\circ} \mathrm{C}\).