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A hot lump of 47.3 g of iron at an initial temperature of 73.5°C is placed in a 50.0 mL H2O initially at 25°C in allowed three thermal equilibrium what is the final temperature of the iron and water given that the specific heat of iron is 0.449 $ divided by parentheses grams times degrees Celsius parentheses this assuming you know he is lost to the surroundings

User Drch
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Answer:

To solve this problem, you can use the principle of conservation of energy, which states that the heat lost by the hot iron will be equal to the heat gained by the cold water.

The heat transfer equation is given by:

\[ Q_{\text{lost}} = Q_{\text{gained}} \]

The heat lost by the iron can be calculated using the formula:

\[ Q_{\text{lost}} = m \cdot c_{\text{iron}} \cdot \Delta T_{\text{iron}} \]

where:

- \( m \) is the mass of the iron (47.3 g),

- \( c_{\text{iron}} \) is the specific heat of iron (0.449 J/(g·°C)),

- \( \Delta T_{\text{iron}} \) is the change in temperature of the iron.

The heat gained by the water can be calculated using the formula:

\[ Q_{\text{gained}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T_{\text{water}} \]

where:

- \( m_{\text{water}} \) is the mass of the water,

- \( c_{\text{water}} \) is the specific heat of water (4.18 J/(g·°C)),

- \( \Delta T_{\text{water}} \) is the change in temperature of the water.

Since the system reaches thermal equilibrium, the final temperature (\( T_{\text{final}} \)) will be the same for both the iron and the water.

Now, let's calculate the heat gained and lost:

\[ m \cdot c_{\text{iron}} \cdot \Delta T_{\text{iron}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T_{\text{water}} \]

We can substitute the values and solve for \( \Delta T_{\text{iron}} \) or \( \Delta T_{\text{water}} \). Once we have one of these values, we can use it to find the final temperature \( T_{\text{final}} \) using the equation:

\[ T_{\text{final}} = T_{\text{initial, iron}} - \Delta T_{\text{iron}} \]

or

\[ T_{\text{final}} = T_{\text{initial, water}} + \Delta T_{\text{water}} \]

Given that the initial temperature of the iron (\( T_{\text{initial, iron}} \)) is 73.5°C and the initial temperature of the water (\( T_{\text{initial, water}} \)) is 25°C, you can now solve for \( T_{\text{final}} \).

User Ideasthete
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