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A 62.6-gram piece of heated limestone is placed into 75.0 grams of water at 23.1°C. The limestone and the water come to a final temperature of 51.9°C. The specific heat capacity of water is 4.186 joules/gram degree Celsius, and the specific heat capacity of limestone is 0.921 joules/gram degree Celsius.

What was the initial temperature of the limestone? Express your answer to three significant figures.

Answer :

m₁ = mass of water = 75 g

T₁ = initial temperature of water = 23.1 °C

c₁ = specific heat of water = 4.186 J/g°C


m₂ = mass of limestone = 62.6 g

T₂ = initial temperature of limestone = ?

c₂ = specific heat of limestone = 0.921 J/g°C


T = equilibrium temperature = 51.9 °C

using conservation of heat

Heat lost by limestone = heat gained by water

m₂c₂(T₂ - T) = m₁c₁(T - T₁)

inserting the values

(62.6) (0.921) (T₂ - 51.9) = (75) (4.186) (51.9 - 23.1)

T₂ = 208.73 °C

in three significant figures

T₂ = 209 °C

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Rewritten by : Barada

Final answer:

To find the specific heat capacity of the metal, we can use the 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. Substituting the given values and solving, we find that the specific heat capacity of the metal is approximately 0.252 J/g °C.

Explanation:

To find the specific heat capacity of the metal, we can use the 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. In this case, we know the initial and final temperatures of the metal, as well as the mass of the water, so we can plug these values into the equation. Rearranging the equation to solve for c, we have: c = q / (mΔT). Substituting the values and solving, we find that the specific heat capacity of the metal is approximately 0.252 J/g °C.

Learn more about specific heat capacity here:

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