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

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