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A 0.0591 kg ingot of metal is heated to 197 degrees C and then dropped into a beaker containing 0.386 kg of water initially at 23 degrees C. If the final equilibrium temperature of the system is 25.4 degrees C, find the specific heat of the metal. The specific heat of water is 4186 J/kg degree C. Provide the answer in units of J/kg degree C.

Answer :

To determine the specific heat of the metal, use the conservation of energy principle, equating the heat lost by the metal to the heat gained by the water, and solve for the specific heat of the metal using the known masses and temperatures.

The specific heat of the metal can be calculated using the principle of conservation of energy, which states that the heat lost by the metal must equal the heat gained by the water. We can express this principle through the formula:

  • Qmetal = -Qwater

Where Q is the heat transferred, given by the product of mass (m), specific heat capacity (c), and the change in temperature (ΔT). Thus, the calculation is:

  • mmetalcmetalΔTmetal = -mwatercwaterΔTwater

Plugging in the known values and solving for cmetal:

  • (0.0591 kg)(cmetal)(25.4 °C - 197 °C) = -(0.386 kg)(4186 J/kg°C)(25.4 °C - 23 °C)

Now we can solve for cmetal:

  • cmetal = (0.386 kg * 4186 J/kg°C * 2.4 °C) / (0.0591 kg * -171.6 °C)

After calculating, we find the specific heat of the metal.

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Rewritten by : Barada