A 40.0g sample of iron, which has a specific heat capacity of 0.449·J·g−1°C−1, is dropped into an insulated container containing 200.0g of water at 70.0°C and a constant pressure of 1atm. The initial temperature of the iron is 8.6°C.Assuming no heat is absorbed from or by the container, or the surroundings, calculate the equilibrium temperature of the water. Be sure your answer has the correct number of significant digits.
Question
A 40.0g sample of iron, which has a specific heat capacity of 0.449·J·g−1°C−1, is dropped into an insulated container containing 200.0g of water at 70.0°C and a constant pressure of 1atm. The initial temperature of the iron is 8.6°C.Assuming no heat is absorbed from or by the container, or the surroundings, calculate the equilibrium temperature of the water. Be sure your answer has the correct number of significant digits.
Solution
To solve this problem, we need to use the principle of conservation of energy, which states that energy cannot be created or destroyed, only transferred. In this case, the heat lost by the water will be equal to the heat gained by the iron.
The formula for heat transfer is Q = mcΔT, where Q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
First, let's calculate the heat lost by the water:
Q_water = m_water * c_water * (T_initial_water - T_final)
We don't know the final temperature yet, so let's just call it T_final for now. The mass of the water is 200.0g, the specific heat capacity of water is 4.18 J/g°C, and the initial temperature is 70.0°C.
Q_water = 200.0g * 4.18 J/g°C * (70.0°C - T_final)
Next, let's calculate the heat gained by the iron:
Q_iron = m_iron * c_iron * (T_final - T_initial_iron)
The mass of the iron is 40.0g, the specific heat capacity is 0.449 J/g°C, and the initial temperature is 8.6°C.
Q_iron = 40.0g * 0.449 J/g°C * (T_final - 8.6°C)
Since the heat lost by the water is equal to the heat gained by the iron, we can set these two equations equal to each other and solve for T_final:
200.0g * 4.18 J/g°C * (70.0°C - T_final) = 40.0g * 0.449 J/g°C * (T_final - 8.6°C)
Solving this equation gives T_final ≈ 63.9°C. Therefore, the equilibrium temperature of the water is approximately 63.9°C.
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