The conservation of energy in calorimetry we can find the amount of water to transform gold from the liquid state to the so
m = 3.48 10-3 kg = 0.00348 kg
To find
Calorimetry analyzes the change in temperature of a body when there is a heat transfer with another body in contact, this is expressed by the related
Q = m c_e ΔT
Where Q is the thermal energy, m is the mass, c_e the specific heat and ΔT is the temperature variation
The transformation of the state of matter liquid solid and liquid vapor occurs with changes in temperature, since the energy supplied is used to change the separation of the molecules and change state
Q = m L
Where m is the mass and L the latent heat
In this case, the coldest body is the water that, when in contact with the gold, undergoes the following processes
• It is heated from the initial temperature (T₀ = 23ºC) to the evaporation temperature (T_f = 100ºC), this process is described by the expression
Q₁ = m₁ c_{e 1} (T_f- T₀)
Q₂ = m₁ L₁
Where L₁ is the latent heat of evaporation, which for water
L₁ = 2.26 10⁶ J/kg
Consequently the heat absorbed by water is
Q_{abs} = Q₁ + Q₂
Q_{abs} = m₁ c_{e 1} (T_f- T₀) + m₁ L₁
Gold is the hottest body and has a process of change of state from the liquid state to the solid state.
Q_{give } = m₂ L₂
L₂ = 6.44 10⁴ J / kg
If we assume that there are no losses with the environment
Q_{abs} = Q_{give}
m₁ c_{e 1} (T_f- T₀) + m₁ L₁ = m₂ L₂
m₁ = [tex]\frac{m_2 \ L_2}{ c_{e \1} (T_f - T_o) + L_1 }[/tex]
m₁ = [tex]\frac{0.140 \ 6.44 \ 10^4 }{4186 (100-23) + 2.26 \ 10^6}[/tex]
m₁ = 0.9016 104 / 2.592 10⁶
m₁= 3.48 10⁻³ kg
Using the conservation of energy in calorimetry we can find the amount of water to transform gold from the liquid state to the solid state is
m = 3.48 10⁻³ kg
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