a 200g aluminum container has 500g of water with an initial temperature of 22 degree celsius. The following heated pieces of metal are dropped into water at the same time as follows: 300g piece of aluminum heated at 100 degree celsius; a 50g piece of copper heated at 80 degree celsius and a 40g piece of steel heated at 120 degree celsius. What will be the final temperature of the mixture?

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To find the final temperature of the mixture, we can use the principle of conservation of energy.

First, let's calculate the energy gained by each metal:

For the aluminum piece of 300g:
Energy gained = mass × specific heat capacity × change in temperature
= 300g × 0.897 J/g°C × (final temperature - 100°C)

For the copper piece of 50g:
Energy gained = 50g × 0.385 J/g°C × (final temperature - 80°C)

For the steel piece of 40g:
Energy gained = 40g × 0.51 J/g°C × (final temperature - 120°C)

Next, let's calculate the energy lost by the water:

Energy lost = mass × specific heat capacity × change in temperature
= 500g × 4.186 J/g°C × (final temperature - 22°C)

Since energy lost = energy gained (assuming no heat is lost to the surroundings), we can set up the equation:

500g × 4.186 J/g°C × (final temperature - 22°C) = (300g × 0.897 J/g°C × (final temperature - 100°C)) + (50g × 0.385 J/g°C × (final temperature - 80°C)) + (40g × 0.51 J/g°C × (final temperature - 120°C))

Simplifying the equation, we get:

2093(final temperature - 22°C) = (269.1(final temperature - 100°C)) + (19.25(final temperature - 80°C)) + (20.4(final temperature - 120°C))

Solve this equation to find the final temperature of the mixture.

To find the final temperature of the mixture, we can use the principle of conservation of energy. The heat gained by the water and the metals will be equal to the heat lost by the hot pieces of metal.

First, let's calculate the heat gained by the water. We can use the specific heat capacity formula:

Q = mcΔT

Where:
Q = heat gained by the water
m = mass of the water
c = specific heat capacity of water (4.186 J/g°C)
ΔT = change in temperature of the water

Given:
Mass of water (m_water) = 500g
Initial temperature of water (T_initial_water) = 22°C

Using the formula:

Q = m_water * c * ΔT

Q = 500g * 4.186 J/g°C * ΔT

Next, we calculate the heat lost by each piece of metal. We can use the same formula as above to calculate the heat lost for each metal piece. Then, we sum up the heat lost by all the metal pieces.

For the aluminum piece:

Mass of aluminum (m_aluminum) = 300g
Initial temperature of aluminum (T_initial_aluminum) = 100°C

Q_aluminum = m_aluminum * c_aluminum * ΔT_aluminum

For the copper piece:

Mass of copper (m_copper) = 50g
Initial temperature of copper (T_initial_copper) = 80°C

Q_copper = m_copper * c_copper * ΔT_copper

For the steel piece:

Mass of steel (m_steel) = 40g
Initial temperature of steel (T_initial_steel) = 120°C

Q_steel = m_steel * c_steel * ΔT_steel

Finally, the total heat gained by the water will be equal to the total heat lost by the metal pieces. Therefore:

Q_water = Q_aluminum + Q_copper + Q_steel

Now let's solve for the final temperature of the mixture.

Rearranging the equation:

ΔT = (Q_water) / (m_water * c_water)

Substituting the values:

ΔT = (Q_aluminum + Q_copper + Q_steel) / (m_water * c_water)

Using the specific heat capacity values:

c_water = 4.186 J/g°C
c_aluminum = 0.897 J/g°C
c_copper = 0.385 J/g°C
c_steel = 0.466 J/g°C

Plug in the values and calculate ΔT:

ΔT = ((m_aluminum * c_aluminum * ΔT_aluminum) + (m_copper * c_copper * ΔT_copper) + (m_steel * c_steel * ΔT_steel)) / (m_water * c_water)

ΔT = ((300g * 0.897 J/g°C * (100°C - T_initial_water)) + (50g * 0.385 J/g°C * (80°C - T_initial_water)) + (40g * 0.466 J/g°C * (120°C - T_initial_water))) / (500g * 4.186 J/g°C)

Solving this equation will give us the change in temperature (ΔT) of the water. To find the final temperature, we add this change in temperature to the initial temperature of the water:

Final temperature = T_initial_water + ΔT