To analyze the heat transfer that occurs between the aluminum block and the water, we can use the principle of conservation of energy. Heat lost by the aluminum block will be equal to the heat gained by the water until thermal equilibrium is reached.
Given:
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Mass of water (\(m_{water}\)) = 200 mL = 200 g (since the density of water is approximately 1 g/mL)
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Specific heat capacity of water (\(c_{water}\)) = 4.18 J/g°C
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Initial temperature of water (\(T_{i,water}\)) = 20°C
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Mass of aluminum block (\(m_{Al}\)) = (depends on its volume and density)
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Specific heat capacity of aluminum (\(c_{Al}\)) = 0.897 J/g°C
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Initial temperature of aluminum block (\(T_{i,Al}\)) = 300°C
Heat transfer calculations:
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Heat gained by water: \[ Q_{water} = m_{water} \cdot c_{water} \cdot (T_f - T_{i,water}) \] where \(T_f\) is the final equilibrium temperature.
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Heat lost by aluminum: \[ Q_{Al} = m_{Al} \cdot c_{Al} \cdot (T_{i,Al} - T_f) \]
Setting the heat gained equal to the heat lost:
\[ m_{water} \cdot c_{water} \cdot (T_f - T_{i,water}) = m_{Al} \cdot c_{Al} \cdot (T_{i,Al} - T_f) \]
At this point, you would need additional information about the mass of the aluminum block or its dimensions to solve for the specific quantities. However, the general concept remains that the aluminum will lose heat until its temperature decreases to the equilibrium temperature, while the water will gain heat until its temperature increases to that same equilibrium temperature.
If you provide the mass of the aluminum block, we can calculate the equilibrium temperature and the specific heat transfer.
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