Respuesta :
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
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