QuestionJuly 25, 2025

25g of a metal solid transfers 190.8 j of heat to its surroundings and in the process goes from 46^circ C to 29^circ C What is the identity of the solid? Al Ag Pb Fe

25g of a metal solid transfers 190.8 j of heat to its surroundings and in the process goes from 46^circ C to 29^circ C What is the identity of the solid? Al Ag Pb Fe
25g of a metal solid transfers 190.8 j of heat to its surroundings and in the process goes from 46^circ C to 29^circ C What is the identity of the solid?
Al
Ag
Pb
Fe

Solution
4.7(253 votes)

Answer

Fe (Iron) Explanation 1. Calculate the change in temperature \Delta T = 46^{\circ}C - 29^{\circ}C = 17^{\circ}C 2. Use the formula for specific heat capacity **q = mc\Delta T** where q = 190.8 \, \text{J}, m = 25 \, \text{g}, and \Delta T = 17^{\circ}C. 3. Solve for specific heat capacity (c) c = \frac{q}{m\Delta T} = \frac{190.8}{25 \times 17} = 0.45 \, \text{J/g}^{\circ}\text{C} 4. Identify the metal Compare c = 0.45 \, \text{J/g}^{\circ}\text{C} with known values: - Al: 0.897 \, \text{J/g}^{\circ}\text{C} - Ag: 0.235 \, \text{J/g}^{\circ}\text{C} - Pb: 0.128 \, \text{J/g}^{\circ}\text{C} - Fe: 0.450 \, \text{J/g}^{\circ}\text{C}

Explanation

1. Calculate the change in temperature<br /> $\Delta T = 46^{\circ}C - 29^{\circ}C = 17^{\circ}C$<br />2. Use the formula for specific heat capacity<br /> **$q = mc\Delta T$** where $q = 190.8 \, \text{J}$, $m = 25 \, \text{g}$, and $\Delta T = 17^{\circ}C$.<br />3. Solve for specific heat capacity ($c$)<br /> $c = \frac{q}{m\Delta T} = \frac{190.8}{25 \times 17} = 0.45 \, \text{J/g}^{\circ}\text{C}$<br />4. Identify the metal<br /> Compare $c = 0.45 \, \text{J/g}^{\circ}\text{C}$ with known values: <br />- Al: $0.897 \, \text{J/g}^{\circ}\text{C}$<br />- Ag: $0.235 \, \text{J/g}^{\circ}\text{C}$<br />- Pb: $0.128 \, \text{J/g}^{\circ}\text{C}$<br />- Fe: $0.450 \, \text{J/g}^{\circ}\text{C}$
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