A mass of 45.0g of molten silver at its melting point is poured into 750mL of calorimeter water at 25.0C. The final temperature reached by both is 30.0C. If the specific heat capacity of silver is 0.235 kJ/kgC, calculate the molar heat of solidification of silver.
heat produced by solidification + heat lost by Ag cooling + heat gained by calorimeter water = 0
[mass Ag x Hvap] + [mass Ag x specific heat Ag x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
That first terms of [mass Ag x Hvap] should be a negative term. I omitted the - sign in front of it.
21 100
To calculate the molar heat of solidification of silver, we need to use the equation:
q = m × c × ΔT
Where:
q = heat energy
m = mass
c = specific heat capacity
ΔT = change in temperature
First, we need to calculate the heat energy gained by the water and the heat energy lost by the silver. Since energy is conserved, the heat lost by the silver is equal to the heat gained by the water.
The heat gained by the water can be calculated using the equation:
q_water = m_water × c_water × ΔT_water
Where:
m_water = mass of water
c_water = specific heat capacity of water
ΔT_water = change in temperature of water
In this case, the mass of the water is given as 750 mL, which we need to convert to grams. The density of water is approximately 1 g/mL, so the mass of the water is:
m_water = 750 mL × 1 g/mL = 750 g
The specific heat capacity of water is approximately 4.18 J/g°C. However, since the specific heat capacity of silver is given in kJ/kg°C, we need to convert it to J/g°C:
c_silver = 0.235 kJ/kg°C × 1000 J/1 kJ × 1 kg/1000 g = 0.235 J/g°C
Substituting the given values into the equation:
q_water = 750 g × 4.18 J/g°C × (30.0°C - 25.0°C) = 15750 J
Since the heat lost by the silver is equal to the heat gained by the water, we have:
q_silver = -q_water = -15750 J
Now, we need to calculate the mass of silver. We know that the mass of the silver is 45.0 g.
Using the equation for heat energy:
q_silver = m_silver × c_silver × ΔT_silver
Substituting the given values:
-15750 J = 45.0 g × 0.235 J/g°C × (30.0°C - T_f)
Where T_f is the final temperature of the silver and water mixture.
Now, we can solve for T_f:
T_f = 30.0°C - (-15750 J) / (45.0 g × 0.235 J/g°C) = 30.0°C + 1422.46°C ≈ 1452.46°C
Since the final temperature reached by the mixture is 30.0°C, which is below the melting point of silver, it means that the silver did not fully solidify. Therefore, we cannot calculate the molar heat of solidification of silver with the given data.