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6 March, 06:06

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

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  1. 6 March, 09:33
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    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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