A 5.00-g sample of aluminum pellets (specific heat capacity = 0.89 J/°C · g) and a 10.00-g sample of iron pellets (specific heat capacity = 0.45 J/°C · g) are heated to 100.0°C. The mixture of hot iron and aluminum is then dropped into 91.9 g of water at 23.7°C. Calculate the final temperature of the metal and water mixture, assuming no heat loss to the surroundings.

heat lost by Al + heat lost by Fe + heat absorbed by water = 0

[mass Al x specific heat Al x (Tfinal-Tinitial)] + [mass Fe x specific heat Fe x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0. Solve for Tfinal.

To solve this problem, we can use the principle of conservation of energy. The heat gained by the water and the metal must equal the heat lost by the metal. The equation for heat transfer is:

Q = m * c * ΔT

Where Q is the amount of heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.

First, let's calculate the amount of heat gained by the water:

Q_water = m_water * c_water * ΔT_water

The mass of water is given as 91.9 g, the specific heat capacity of water is approximately 4.18 J/°C·g, and the change in temperature is the difference between the final temperature and the initial temperature, which is:

ΔT_water = final temperature - initial temperature = final temperature - 23.7°C

Next, let's calculate the amount of heat lost by the metal:

Q_metal = (m_aluminum * c_aluminum * ΔT_metal) + (m_iron * c_iron * ΔT_metal)

The mass of aluminum is given as 5.00 g, the specific heat capacity of aluminum is 0.89 J/°C·g, the mass of iron is given as 10.00 g, the specific heat capacity of iron is 0.45 J/°C·g, and the change in temperature is the difference between the initial temperature (100.0°C) and the final temperature.

Since there is no heat lost to the surroundings, we can set Q_water equal to Q_metal and solve for the final temperature.

m_water * c_water * ΔT_water = (m_aluminum * c_aluminum * ΔT_metal) + (m_iron * c_iron * ΔT_metal)

Now we can substitute the given values into the equation and solve for the unknown final temperature.

To solve this problem, we can use the principle of conservation of energy, which states that the heat gained by the substances will equal the heat lost.

First, let's calculate the heat gained by the aluminum and iron pellets:
The heat gained by the aluminum pellets can be calculated using the formula:
Q = m * c * ΔT
where Q is the heat gained, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

For aluminum:
Q_aluminum = m_aluminum * c_aluminum * ΔT
= 5.00 g * 0.89 J/°C · g * (100.0°C - 23.7°C)
= 5.00 g * 0.89 J/°C · g * 76.3°C
= 340.27 J

Similarly, the heat gained by the iron pellets can be calculated as:
Q_iron = m_iron * c_iron * ΔT
= 10.00 g * 0.45 J/°C · g * (100.0°C - 23.7°C)
= 10.00 g * 0.45 J/°C · g * 76.3°C
= 343.35 J

Next, let's calculate the heat lost by the aluminum and iron pellets when they are added to the water. We can use the same formula:
Q_lost = m_water * c_water * ΔT
where Q_lost is the heat lost, m_water is the mass of water, c_water is its specific heat capacity, and ΔT is the change in temperature.

The heat lost by the aluminum and iron pellets will be the same, as they cool down to the final temperature of the mixture.

Now, we need to determine the final temperature of the mixture. Let's assume it is Tf.

The heat lost by the aluminum and iron pellets is equal to the heat gained by the water:
Q_lost = Q_water

Plugging in the values, we get:
m_water * c_water * ΔT = Q_water
91.9 g * 4.186 J/g·°C * (Tf - 23.7°C) = 340.27 J + 343.35 J
382.80 g·°C * (Tf - 23.7°C) = 683.62 J

Simplifying the equation:
382.80 Tf - 382.80 * 23.7°C = 683.62 J
382.8 Tf - 9082.76 = 683.62
382.8 Tf = 9766.38
Tf = 9766.38 / 382.8
Tf = 25.49°C

Therefore, the final temperature of the metal and water mixture is approximately 25.49°C.