3. An irregular lump of an unknown metal has a measured density of \( 2.97 \mathrm{~g} / \mathrm{mL} \). The metal is heated to a temperature of \( 173^{\circ} \mathrm{C} \) and placed in a graduated cylinder filled with 25.0 mL of water at \( 25.0^{\circ} \mathrm{C} \). After the system has reached thermal equilibrium, the volume in the cylinder is read at 34.0 mL , and the temperature is recorded as \( 40.6^{\circ} \mathrm{C} \). What is the specific heat of the unknown metal sample? Assume no heat is lost to the surroundings. What is the metal used in the experiment? Specific heat of water \( =4.184 \mathrm{~J} / \mathrm{g} \mathrm{C} \) Specific heat of granite \( =0.79 \mathrm{~J} / \mathrm{g} \mathrm{C} \) Specific heat of copper \( =0.38 \mathrm{~J} / \mathrm{g} \mathrm{C} \) Specific heat of iron \( =0.45 \mathrm{~J} / \mathrm{g} \mathrm{C} \) Specific heat of lead \( =0.13 \mathrm{~J} / \mathrm{g} \mathrm{C} \)
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To find the specific heat of the unknown metal, we need to use the principle of conservation of energy. The heat lost by the metal will be equal to the heat gained by the water. First, calculate the mass of water: - Mass of water (m_water) = Volume × Density = 25.0 mL × 1 g/mL = 25.0 g. - Change in temperature of water (ΔT_water) = Final temperature - Initial temperature = 40.6 °C - 25.0 °C = 15.6 °C. Now, the heat gained by the water can be calculated as: \[ Q_{water} = m_{water} × c_{water} × \Delta T_{water} = 25.0 \, g × 4.184 \, \frac{J}{g \, °C} × 15.6 \, °C \] \[ Q_{water} = 1623 J \] Assuming no heat is lost to the surroundings, this is equal to the heat lost by the metal: \[ Q_{metal} = -Q_{water} \] Let the mass of the metal (m_metal) be determined from its density: - Volume of the metal = Final volume - Initial volume of water = 34.0 mL - 25.0 mL = 9.0 mL. - Mass of metal (m_metal) = Density × Volume = 2.97 g/mL × 9.0 mL = 26.73 g. The change in temperature of the metal (ΔT_metal) is: - ΔT_metal = Initial temperature of metal - Final temperature = 173 °C - 40.6 °C = 132.4 °C. Now we use the heat lost by the metal: \[ Q_{metal} = m_{metal} × c_{metal} × ΔT_{metal} = -1623 J \] \[ 26.73 g × c_{metal} × 132.4 °C = -1623 J \] Finally, we can solve for the specific heat of the metal: \[ c_{metal} = \frac{-1623 J}{26.73 g × 132.4 °C} \approx 0.45 \, \frac{J}{g \, °C} \] Given this specific heat, the metal could be iron, as it matches closely with the known specific heats provided. Did you know that the specific heat of a substance tells us how much heat it can absorb per unit mass? It's like a measure of how well a material can hang onto its heat—think of it as its "thermal personality!" The higher the specific heat, the less temperature change you'll see with the addition of heat. If you're diving into the world of specific heats, consider looking into thermal conductivity too! It's related, as it measures how well a substance conducts heat. You could explore materials used in cooking, like stainless steel versus cast iron, and see how their heat properties influence cooking performance. Hot stuff, literally!
