Question:

A geyser heats water flowing at the rate of 3.0 litres per minute from 27oC to 77oC. If the geyser operates on a gas burner. What is the rate of consumption of the fuel if its heat of combustion is 4.0×10^4 J/g?

Answer:

1. Determine the amount of heat energy required to heat the water: Q = m x c x ΔT Q = 3.0 L x 4.18 J/g°C x (77°C - 27°C) Q = 5.7 × 10^5 J

2. Determine the rate of fuel consumption: Rate = (Heat energy required) / (Heat of combustion) Rate = 5.7 × 10^5 J / 4.0 × 10^4 J/g Rate = 1.425 g/min

Question:

A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume?

Answer:

1. Calculate the original volume of the gas: 3 moles of hydrogen at standard temperature and pressure have a volume of 24 L.

2. Calculate the new volume of the gas after compression: The new volume of the gas is half of the original volume, which is 12 L.

3. Calculate the new pressure of the gas after compression: The new pressure of the gas is twice the original pressure, which is 2 atmospheres.

Question:

ProcessIn changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case? (Take 1 cal = 4.19 J)

Answer:

Answer:

1. Convert 9.35 cal to joules: 9.35 cal * 4.19 J/cal = 38.945 J

2. Subtract the net heat absorbed from the amount of work done: 22.3 J - 38.945 J = -16.645 J

3. The net work done by the system is -16.645 J.

Question:

Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is now opened. Answer the following : (a) What is the final pressure of the gas in A and B ? (b) What is the change in internal energy of the gas? (c) What is the change in the temperature of the gas? (d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P-V-T surface ?

Answer:

(a) The final pressure of the gas in A and B will be the same and equal to the standard temperature and pressure.

(b) The change in internal energy of the gas will be zero, as it is a thermally insulated system.

(c) The change in temperature of the gas will be zero, as the system is thermally insulated.

(d) The intermediate states of the system will not lie on its P-V-T surface, as the pressure and temperature remain constant throughout the process.

Question:

What amount of heat must be supplied to 2.0×10^−2kg of nitrogen (at room temperature) to raise its temperature by 45oC at constant pressure? (Molecular mass of N2​=28:R=8.3mol^−1K^−1.)

Answer:

Step 1: Calculate the amount of heat required (Q).

Q = m × c × ΔT

Step 2: Substitute the given values in the equation.

Q = 2.0 × 10^−2 kg × (8.3 mol^−1K^−1 × 28) × 45oC

Step 3: Calculate the amount of heat required.

Q = 8.1 × 10^2 J

Question:

Explain why (a) Two bodies at different temperatures T1​ and T2​ if brought in thermal contact do not necessarily settle to the mean temperature (T1​ +T2​)/2 (b) The coolant in a chemical or a nuclear plant (i.e. the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat (c) Air pressure in a car tyre increases during driving (d) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude

Answer:

(a) Two bodies at different temperatures T1 and T2 if brought in thermal contact do not necessarily settle to the mean temperature (T1 +T2)/2 because the rate of heat transfer between them depends on the thermal conductivity of the material between them, which can be different. Heat transfer between two objects is also affected by the surface area of the objects, the temperature difference between them, and the time they are in contact.

(b) The coolant in a chemical or a nuclear plant (i.e. the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat because it needs to absorb a lot of heat from the different parts of the plant without changing its temperature significantly. This is important to ensure that the temperature of the different parts of the plant stays at a safe level.

(c) Air pressure in a car tyre increases during driving because the air inside the tyre gets compressed due to the increased pressure from the road. This increased pressure causes the air inside the tyre to heat up, which increases the air pressure inside the tyre.

(d) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude because the presence of water bodies around the harbour town moderates the temperature, making it more temperate than the desert town. This is because water has a higher heat capacity than land, which means it absorbs and releases heat more slowly than land. This means the temperature in the harbour town is less likely to fluctuate drastically.

Question:

A refrigerator is to maintain eatables kept inside at 9oC. If room temperature is 36oC, calculate the coefficient of performance.

Answer:

Step 1: Calculate the temperature difference between the inside of the refrigerator and the room temperature.

Temperature difference = 36°C - 9°C = 27°C

Step 2: Calculate the coefficient of performance.

Coefficient of Performance = Temperature Difference / Room Temperature

Coefficient of Performance = 27°C / 36°C = 0.75

Question:

An electric heater supplies heat to a system at a rate of 100W. If system performs work at a rate of 75 joules per second. At what rate is the internal energy increasing?

Answer:

1. The rate at which the internal energy of the system is increasing is equal to the rate of heat supplied minus the rate of work performed.

2. Therefore, the rate of increase of internal energy is 100W (rate of heat supplied) - 75 joules per second (rate of work performed) = 25W.

Question:

A steam engine delivers 5.4×10^8 J of work per minute and services 3.6×10^9 J of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute?

Answer:

Efficiency = (Work delivered per minute) / (Heat serviced per minute)

Efficiency = 5.4×10^8 J / 3.6×10^9 J

Efficiency = 0.15

Heat Wasted per minute = (Heat serviced per minute) - (Work delivered per minute)

Heat Wasted per minute = 3.6×10^9 J - 5.4×10^8 J

Heat Wasted per minute = 3.05×10^9 J