I want to see if i got my answer correct for #1 but I need help on #2
1.A 150g aluminum pan containing 500g of water is heated on a stove for 5 minutes. The temperature goes from 23.5C to 48.9C. How much heat was absorbed. (s for Al is .900j/g C, s for H2O is 4.184 j/g C) I got 3429J AL and 53136.8J H2O heat absorbed?
2. When a 1.000 g sample of the rocket fuel hydrazine, N2H4, is burned in a bomb calorimeter which contains 1200g of water, the temperature rises from 24.62C to 28.16C. If the C for the bomb is 840 J/C
Calculate: q reaction for combustion of a one-gram sample
and
q reaction for combustion of one mole of hydrazine in the bomb calorimeter
To calculate the heat absorbed in each scenario, you can use the formula: q = m * s * ΔT, where q represents heat absorbed, m represents mass, s represents specific heat, and ΔT represents the change in temperature.
1. For the aluminum pan and water scenario:
Aluminum:
q_Al = m_Al * s_Al * ΔT_Al
q_Al = 150g * 0.900 J/g °C * (48.9°C - 23.5°C)
q_Al = 150g * 0.900 J/g °C * 25.4°C
q_Al = 3423 J (rounded to the nearest whole number)
Water:
q_H2O = m_H2O * s_H2O * ΔT_H2O
q_H2O = 500g * 4.184 J/g °C * (48.9°C - 23.5°C)
q_H2O = 500g * 4.184 J/g °C * 25.4°C
q_H2O = 53140 J (rounded to the nearest whole number)
So your calculation for heat absorbed in the aluminum pan and water scenario is correct.
2. For the combustion of hydrazine in a bomb calorimeter:
First, calculate the heat absorbed for the one-gram sample:
q_1gram = m_water * C_bomb * ΔT
q_1gram = 1200g * 840 J/°C * (28.16°C - 24.62°C)
q_1gram = 1200g * 840 J/°C * 3.54°C
q_1gram = 3546,240 J (rounded)
Next, calculate the heat absorbed for one mole of hydrazine:
The molar mass of hydrazine (N2H4) is 32.04 g/mol.
1 mole of hydrazine is equal to 32.04 grams.
q_1mole = (q_1gram / 1g) * (32.04g / 1mol)
q_1mole = 3546,240 J * (32.04g / 1g)
q_1mole = 113,649,254 J (rounded)
So the calculations for both the one-gram sample and one mole of hydrazine are correct.
Please note that these calculations assume no heat loss to the surroundings or other factors that may affect the accuracy of the results.
To solve #2, we need to calculate the heat absorbed during the combustion reaction of hydrazine.
First, let's calculate the heat absorbed when a 1.000 g sample of hydrazine is burned in the bomb calorimeter, which contains 1200 g of water.
To calculate the heat absorbed (q) for the combustion of the one-gram sample, we can use the formula:
q = m * C * ΔT
Where:
q is the heat absorbed,
m is the mass of the substance (1.000 g in this case),
C is the heat capacity of the system (in this case, the bomb calorimeter),
ΔT is the change in temperature (Tfinal - Tinitial).
Given:
m = 1.000 g
C = 840 J/°C
ΔT = 28.16°C - 24.62°C = 3.54°C
Substituting the values into the formula:
q = 1.000 g * 840 J/°C * 3.54°C
q = 2968.8 J
Therefore, the heat absorbed during the combustion of a one-gram sample of hydrazine is 2968.8 J.
To calculate the heat absorbed during the combustion of one mole of hydrazine, we need to determine the molar mass of hydrazine (N2H4).
The molecular weight of nitrogen (N) is 14.01 g/mol, and the molecular weight of hydrogen (H) is 1.01 g/mol. Since hydrazine contains 2 nitrogen atoms and 4 hydrogen atoms, the molar mass of hydrazine is:
2 * (14.01 g/mol) + 4 * (1.01 g/mol) = 32.05 g/mol
Now, we can calculate the heat absorbed.
The heat absorbed during the combustion of one mole of hydrazine is equal to the heat absorbed for the one-gram sample multiplied by the molar mass of hydrazine:
q = (2968.8 J) * (32.05 g/mol)
q ≈ 95,004 J/mol
Therefore, the heat absorbed during the combustion of one mole of hydrazine in the bomb calorimeter is approximately 95,004 J/mol.