Respuesta :
mâ = mass of water = 75 g
Tâ = initial temperature of water = 23.1 °C
câ = specific heat of water = 4.186 J/g°C
mâ = mass of limestone = 62.6 g
Tâ = initial temperature of limestone = ?
câ = 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
mâcâ(Tâ - T) = mâcâ(T - Tâ)
inserting the values
(62.6) (0.921) (Tâ - 51.9) = (75) (4.186) (51.9 - 23.1)
Tâ = 208.73 °C
in three significant figures
Tâ = 209 °C
Tâ = initial temperature of water = 23.1 °C
câ = specific heat of water = 4.186 J/g°C
mâ = mass of limestone = 62.6 g
Tâ = initial temperature of limestone = ?
câ = 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
mâcâ(Tâ - T) = mâcâ(T - Tâ)
inserting the values
(62.6) (0.921) (Tâ - 51.9) = (75) (4.186) (51.9 - 23.1)
Tâ = 208.73 °C
in three significant figures
Tâ = 209 °C