If someone could see if I have the right answers, and if I don't maybe give me some hints on what to do to get to the right answerr, please.
1. Calculate the heat required to raise the temperature of 150g of water by 25.0°C. [For pure water specific heat capacity=4.18Jg^-1K^-1]
Answer=15.675kJ ? (using mass*specific heat capacity*change in temp?)
2. The calorimeter constant of a flame calorimeter is 5.83*10^3JK^-1. The complete combustion of 3.20g of methanol CH3Oh raises the temperature by 12.3°C. Calculate the enthalpy change for the combustion of 1.00mol of methanol.
Answer: bit stuck here. not sure what to or how to include all the numbers.
2341.51kJ ? (by finding out the enthalpy change, then the molar mass of methanol and the moles, then using the moles to divide by the enthalpy change?)
#1 is ok.
To calculate the heat required to raise the temperature of water, you can use the formula:
Q = m * c * ΔT
where:
Q is the heat (in joules)
m is the mass of the substance (in grams)
c is the specific heat capacity of the substance (in joules per gram per degree Celsius)
ΔT is the change in temperature (in degrees Celsius)
Let's apply this formula to the first question:
1. Calculate the heat required to raise the temperature of 150g of water by 25.0°C.
Q = (150g) * (4.18J/g°C) * (25.0°C) = 15,675J
You are correct in multiplying the mass (150g) by the specific heat capacity (4.18J/g°C) and the change in temperature (25.0°C). However, the answer is in joules, not kilojoules. So the correct answer would be 15,675J, not 15.675kJ.
Moving on to the second question:
2. Calculate the enthalpy change for the combustion of 1.00mol of methanol.
Using the data provided, you need to find the enthalpy change (ΔH) for the combustion of 3.20g of methanol. Then you can calculate the enthalpy change for 1.00mol of methanol using stoichiometry.
First, calculate the heat released by the combustion of 3.20g of methanol using the formula from question 1:
Q = (3.20g) * (4.18J/g°C) * (12.3°C) = 160.1664J
Now, convert the heat released to kilojoules:
Q = 160.1664J = 0.1601664kJ
Next, use the flame calorimeter constant (5.83 x 10^3 J/K) to calculate the enthalpy change for 3.20g of methanol:
ΔH = Q / (constant * ΔT) = (0.1601664kJ) / [(5.83 x 10^3 J/K) * (12.3°C)] = 0.0026301 kJ/K
Now, we need to calculate the enthalpy change for 1.00mol of methanol. To do this, we need to know the molar mass of methanol (CH3OH), which is 32.04g/mol.
The number of moles of methanol (n) can be calculated using the mass given:
n = mass / molar mass = 3.20g / 32.04g/mol = 0.10mol
Finally, calculate the enthalpy change for 1.00mol of methanol:
ΔH = (enthalpy change for 3.20g) * (1.00mol / 0.10mol) = (0.0026301kJ/K) * 10 = 0.026301kJ
So, the correct answer for the enthalpy change for the combustion of 1.00mol of methanol is approximately 0.0263kJ.