what is the molar enthalpy change when 147 g of water cools from 90.0 degrees C to 17.0 degrees C
q = mass H2O x specific heat H2O x (Tfinal-Tinitial).
I would convert gH2O to mols (147/18).
Use specific heat in J/mol = about 75.
To calculate the molar enthalpy change, we need to know the specific heat capacity of water and the molar mass of water.
The specific heat capacity of water is given as 4.18 J/g°C, which means it takes 4.18 Joules of energy to raise the temperature of 1 gram of water by 1 degree Celsius.
First, we need to calculate the mass of water in moles using the molar mass of water. The molar mass of water (H2O) is approximately 18.015 g/mol.
Mass of water = 147 g
Molar mass of water = 18.015 g/mol
Number of moles of water = mass of water / molar mass of water
Number of moles of water = 147 g / 18.015 g/mol
Next, we calculate the temperature change.
Initial temperature, T1 = 90.0°C
Final temperature, T2 = 17.0°C
Temperature change ΔT = T2 - T1
Temperature change ΔT = 17.0°C - 90.0°C
Now we can calculate the molar enthalpy change using the formula:
Enthalpy change (ΔH) = (number of moles of water) x (specific heat capacity of water) x (temperature change)
Substituting the values we calculated:
ΔH = (147 g / 18.015 g/mol) x (4.18 J/g°C) x (-73.0°C)
Now we can solve for ΔH:
ΔH ≈ - 206,084 J/mol
So, the molar enthalpy change when 147 g of water cools from 90.0°C to 17.0°C is approximately -206,084 J/mol.