High School

A 55.0 g piece of metal was heated in boiling water to 99.8 degrees Celsius and then dropped into water in an insulated beaker. There are 225 mL of water (density = 1.00 g/mL) in the beaker, and its temperature before the metal was dropped in was 21.0 degrees Celsius. The final temperature of the metal and the water is 23.1 degrees Celsius. What is the specific heat of the metal?

Answer :

Final answer:

The specific heat of the metal can be found by applying the principle of heat conservation or the principle of calorimetry. The heat lost by the hot metal is equal to the heat gained by the cooler water. Solving this equation will give the specific heat of the metal.

Explanation:

To solve this problem, we will use the heat conservation equation. As per the conservation of energy, the heat lost by the hot metal should be equal to the heat gained by the cooler water. So, this can be expressed as: (Cmetal)(55.0g)(23.1°C - 99.8°C) = -(4.18 J/g °C)(225g)(23.1°C - 21.0°C). Here, Cmetal is the specific heat of the metal, we will solve this equation for Cmetal, which will give us the specific heat of the metal in terms of J/g °C. Such problems are based on the principle of calorimetry where the energy lost by a hot object equals the energy gained by a cool object.

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