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Identify the specific heat of a 55.0g sample of metal that started at 99.0C when dropped into 350. grams of water at 22.0C. The temperature of the water changed to 22.7C.

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Final answer:

The specific heat of the metal can be calculated using the equation q = mcΔT, where q is the heat transferred, m is the mass of the metal, c is the specific heat capacity of the metal, and ΔT is the change in temperature. By equating the heat transferred to the water and the heat transferred from the metal, we can solve for the specific heat of the metal. Using the provided values, the specific heat of the metal is approximately 0.96 J/g°C.

Step-by-step explanation:

The specific heat of a metal can be calculated using the equation:

q = mcΔT

Where:

  • q is the heat transferred
  • m is the mass of the metal
  • c is the specific heat capacity of the metal
  • ΔT is the change in temperature

In this case, the metal was dropped into water, so we can assume that all the heat transferred to the water.

Using the equation:

  • For the metal: qmetal = mcmetalΔT
  • For the water: qwater = mcwaterΔT

Since the heat transferred to the water is equal to the heat transferred from the metal, we can set up the following equation:

mcmetalΔT = mcwaterΔT

Simplifying the equation:

cmetal = (mwatercwaterΔT)/(mmetalΔT)

Substituting the given values:

cmetal = (350. g)(4.18 J/g°C)(22.7 °C - 22.0 °C)/(55.0 g)(99.0 °C - 22.7 °C)

After evaluating the expression, the specific heat of the metal is approximately 0.96 J/g°C.

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