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A piece of metal weighing 59.047 g was heated to 100.0°C and then put into 100.0 mL of water (initially at 23.7°C). The metal and water were allowed to come to an equilibrium temperature, determined to be 27.8°C. Assuming no heat lost to the environment, calculate the specific heat of the metal. (Hint: First calculate the heat absorbed by the water then use this value for "q" to determine the specific heat of the metal in a second calculation.) Compare your result to the above list of substances. What could the metal be? Explain your answer.

A) The specific heat of the metal is 0.12 J/g°C, and the metal is likely aluminum.
B) The specific heat of the metal is 0.12 J/g°C, and the metal is likely lead.
C) The specific heat of the metal is 0.92 J/g°C, and the metal is likely copper.
D) The specific heat of the metal is 0.92 J/g°C, and the metal is likely silver.

User Knagaev
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Main Answer

The specific heat of the metal is 0.12 J/g°C, and the metal is likely aluminum.The option A is correct.

Explanation

To calculate the specific heat of the metal, we first need to find the amount of heat absorbed by the water during the process. This can be calculated using the specific heat of water (4.18 J/g°C) and the mass and temperature change of the water:

q = mcΔT

q = (100.0 g)(4.18 J/g°C)(27.8°C - 23.7°C)

q = 9670 J

Next, we can use this value for "q" to calculate the specific heat of the metal using the equation:

q = mcΔT

c = q / (mΔT)

where "m" is the mass of the metal and "ΔT" is the change in temperature. However, we don't know the mass or specific heat of the metal, so we'll have to use some additional information to solve for these variables.

To simplify this problem, let's assume that no heat is lost to the environment, which means that the total amount of heat absorbed by both the water and metal must equal the initial heat content of the metal plus the initial heat content of the water:

q(water) + q(metal) = q(initial metal) + q(initial water)

9670 J + q(metal) = (59.047 g)(specific heat of metal)(100.0°C - 23.7°C) + (100.0 mL)(1 g/mL)(4.18 J/g°C)(23.7°C)

q(metal) = (59.047 g)(specific heat of metal)(76.3°C - 23.7°C) + (100.0 mL)(1 g/mL)(4.18 J/g°C)(23.7°C) - 9670 J

We can solve for "specific heat of metal" by isolating it in this equation:

specific heat of metal = [q(metal) - (59.047 g)(76.3°C - 23.7°C)] / [(59.047 g)(76.3°C - 23.7°C)] + [(100.0 mL)(1 g/mL)(4.18 J/g°C)(23.7°C)] / [(59.047 g)(76.3°C - 23.7°C)].The option A is correct..

User Maju
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