1. A system absorbs 1.57E2 kJ of heat and the surroundings do 1.096E2 kJ of work on the system. What is the change in internal energy (in kJ) (ΔU) of the system?
2. The air in an inflated balloon (the system) absorbs 125 J of heat from its proximity to a fire. The balloon expands and does 85 kJ of work. What is the change in internal energy (in kJ) (\DeltaU) for the system? Enter only the numerical value.
3. How much heat is liberated (in kJ) from 250 g of silver when it cools from 86 °C to 26 °C? The heat capacity of silver is 0.235 Jg^{-1} °C^{-1}. Note, "heat liberated" implies that the change in heat is negative. Enter a positive number.
4. When 0.1523 g of liquid pentane (C_5H_{12}) combusts in a bomb calorimeter, the temperature rises from 23.7^{\circ}C to 29.8 ^{\circ}C. What is \DeltaU_{rxn} for the reaction in kJ/mol pentane? The heat capacity of the bomb calorimeter is 5.23 kJ/^{\circ}C.
5. When 0.1625 g of magnesium is burned in a bomb calorimeter that has a heat capacity of 3.03 kJ/^{\circ}C, the temperature increases by 1.252^{\circ}C. How much heat (kJ/mol) is liberated during the burning of the magnesium?
wondering if this is not a test in heat energy, but a test of handling significant digits....or it could be that your instuctor does not care about precision. For example (they all offer similar analysis), question 4:
you have a multiplication of a four sig digit 0.1523 then a two digit (29.8-23.7 )=6.1, then a three digit 5.23. And the proper answer is only accurate to two significant digits. Wondering what the machine will count as correct...
good luck.
1. To find the change in internal energy of the system (ΔU), you can use the first law of thermodynamics equation:
ΔU = Q - W
Where Q is the heat absorbed by the system and W is the work done on the system by the surroundings.
Given that Q = 1.57E2 kJ (157 kJ) and W = -1.096E2 kJ (-109.6 kJ) (negative because work is done on the system), we can substitute these values into the equation:
ΔU = 1.57E2 kJ - (-1.096E2 kJ)
ΔU = 2.666E2 kJ
Therefore, the change in internal energy of the system is 2.666E2 kJ.
2. Following the same equation as in question 1, we have:
ΔU = Q - W
Given that Q = 125 J and W = 85 kJ (convert kJ to J by multiplying by 1000), we can substitute these values into the equation:
ΔU = 125 J - 85 kJ (note that 85 kJ equals 85,000 J)
ΔU = -84,875 J
Therefore, the change in internal energy of the system is -84,875 J.
3. To calculate the heat liberated (ΔQ), you can use the heat capacity equation:
ΔQ = mcΔT
Where m is the mass, c is the heat capacity, and ΔT is the change in temperature.
Given that m = 250 g, c = 0.235 Jg^(-1)°C^(-1), and ΔT = 86 °C - 26 °C (60 °C), we can substitute these values into the equation:
ΔQ = 250 g × 0.235 Jg^(-1)°C^(-1) × 60 °C
ΔQ = 3,525 J
Therefore, the heat liberated from 250 g of silver cooling from 86 °C to 26 °C is 3,525 J.
4. To find ΔU_rxn for the reaction in kJ/mol of pentane, we can use the equation:
ΔU_rxn = ΔU + ΔU_cal
Where ΔU is the change in internal energy of the system and ΔU_cal is the change in internal energy of the calorimeter.
Given that the mass of pentane is 0.1523 g, the specific heat capacity of pentane is 0.258 J/g°C, the change in temperature is 29.8 °C - 23.7 °C (6.1 °C), and the heat capacity of the calorimeter is 5.23 kJ/°C, we can calculate:
ΔU = m × c × ΔT
ΔU = 0.1523 g × 0.258 J/g°C × 6.1 °C
ΔU = 0.2409 J
ΔU_cal = ΔT_cal × C_cal
ΔU_cal = 6.1 °C × 5.23 kJ/°C (convert kJ to J by multiplying by 1000)
ΔU_cal = 31.963 kJ (31,963 J)
ΔU_rxn = 0.2409 J + 31.963 kJ (convert J to kJ by dividing by 1000)
ΔU_rxn = 32.203 kJ (32,203 J)
Therefore, ΔU_rxn for the combustion of liquid pentane is 32.203 kJ/mol pentane.
5. Similar to question 4, we can use the equation:
ΔU_rxn = ΔU + ΔU_cal
Given that the mass of magnesium is 0.1625 g, the change in temperature is 1.252 °C, and the heat capacity of the calorimeter is 3.03 kJ/°C, we can calculate:
ΔU = m × c × ΔT
ΔU = 0.1625 g × 3.03 J/g°C × 1.252 °C
ΔU = 0.61051 J
ΔU_cal = ΔT_cal × C_cal
ΔU_cal = 1.252 °C × 3.03 kJ/°C (convert kJ to J by multiplying by 1000)
ΔU_cal = 3.79756 kJ (3,797.56 J)
ΔU_rxn = 0.61051 J + 3.79756 kJ (convert J to kJ by dividing by 1000)
ΔU_rxn = 4.40807 kJ (4,408.07 J)
Therefore, the heat liberated during the burning of magnesium is 4.40807 kJ/mol magnesium.
1. To find the change in internal energy (ΔU) of the system, you can use the First Law of Thermodynamics, which states that the change in internal energy is equal to the heat added to the system minus the work done by the system:
ΔU = Q - W
Given that the system absorbs 1.57E2 kJ of heat and the surroundings do 1.096E2 kJ of work on the system, you can substitute these values into the equation:
ΔU = 1.57E2 kJ - 1.096E2 kJ
Calculating this, you will find that the change in internal energy of the system is:
ΔU = 0.474E2 kJ
2. Similar to the previous question, you can use the First Law of Thermodynamics to determine the change in internal energy (ΔU) of the system:
ΔU = Q - W
Given that the system absorbs 125 J of heat and does 85 kJ of work, you need to ensure that the units of heat and work are consistent. Convert the 85 kJ of work to J:
85 kJ = 85E3 J
Substituting the values into the equation, you have:
ΔU = 125 J - 85E3 J
Simplifying this, you will find that the change in internal energy of the system is:
ΔU = -84.875E3 J
3. To calculate the heat liberated when a substance cools, you can use the formula:
Q = m * c * ΔT
Where Q is the heat, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Given that you have the mass of 250 g (which is 0.25 kg), the specific heat capacity of silver as 0.235 Jg^(-1)°C^(-1), and the change in temperature from 86°C to 26°C (which is -60°C), you can substitute these values into the formula:
Q = 0.25 kg * 0.235 Jg^(-1)°C^(-1) * (-60°C)
Converting the mass to grams, the equation becomes:
Q = 250 g * 0.235 Jg^(-1)°C^(-1) * (-60°C)
Upon multiplying and simplifying, you will find that the heat liberated is:
Q = -3525 J
Since the question asks for a positive number, the heat liberated is 3525 J.
4. To find the change in internal energy (ΔU) for the reaction in kJ/mol of pentane, you can use the equation:
ΔU = Q - W
Given that the bomb calorimeter's heat capacity is 5.23 kJ/°C, the temperature change is from 23.7°C to 29.8°C (which is ΔT = 29.8°C - 23.7°C = 6.1°C), and the mass of liquid pentane is 0.1523 g, you can calculate the heat (Q) from the equation:
Q = m * C * ΔT
Where m is the mass, C is the heat capacity, and ΔT is the change in temperature.
Converting the mass to moles using the molar mass of pentane (C5H12), you can calculate the number of moles (n) of pentane:
n = mass / molar mass
Given that the molar mass of pentane (C5H12) is 72.15 g/mol, you can calculate the number of moles:
n = 0.1523 g / 72.15 g/mol
Now, substitute the calculated number of moles into the equation for heat (Q):
Q = n * ΔU
Since the temperature change (ΔT) is in °C, you need to convert it to Kelvin (K):
ΔT(K) = ΔT(°C) + 273.15
Calculate the change in internal energy (ΔU) using the equation:
ΔU = Q / n
Finally, convert the change in internal energy from J to kJ:
ΔU(kJ/mol) = ΔU(J/mol) / 1000
5. To determine the amount of heat liberated during the burning of magnesium, you can use the equation:
Q = m * C * ΔT
Given that you have the mass of magnesium as 0.1625 g, the heat capacity of the bomb calorimeter as 3.03 kJ/°C, and the temperature change as 1.252°C, you can substitute these values into the equation:
Q = 0.1625 g * 3.03 kJ/°C * 1.252°C
Simplifying this equation, you will find that the heat liberated is:
Q = 0.6152 kJ