A 0.0421 kg ingot of metal is heated to 187 ∘C and then is dropped into a beaker containing 0.374 kg of water initially at 22 ∘C. If the final equilibrium state of the mixed system is 24.4 ∘C, find the specific heat of the metal. The specific heat of water is 4186 J/kg⋅ ∘C.

Answer in units of J/kg⋅ ∘C.

To find the specific heat of the metal, equate the heat lost by the metal with the heat gained by the water and solve for the specific heat of the metal using the formula q = mcΔT.

Explanation:

In this problem, we are dealing with what is referred to as an equilibrium state in a mixed system. The key point here is that the heat lost by the metal equals the heat gained by the water.

To solve for the specific heat of the metal, we need to apply the heat transfer formula: q = mcΔT, where q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

For the metal, the heat loss can be expressed as q_metal = c_metal x 0.0421 kg x (187 - 24.4) °C. For the water, the heat gain is q_water = 4186 J/kg°C x 0.374 kg x (24.4 - 22) °C.

By setting these two equations equal to each other, we find c_metal = (4186 J/kg°C x 0.374 kg x (24.4 - 22) °C) / (0.0421 kg x (187 - 24.4) °C), which is the solution to the problem.

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