Do the A/C and heater in a car use gas?
The heater is grabbing a bit of heat from your circulating engine coolant (that normally would be dumped into the outside air by your radiator) and, instead, dumping it into your inside air with a miniature radiator inside your dashboard that your cabin fan blows across. Turning on the heater is just opening a valve to let varying flow rates of hot coolant into the small radiator, allowing varying levels of heat to be picked up by the passing cabin air being blown by. By and large, no use of gas, unless, as previously mentioned, you weren’t using the fan until you turned on the heat.The AC is a belt-driven mechanical compressor that compresses and circulates refrigerant through a closed loop between another in-the-dash heat exchanger/mini-radiator and an external heat radiator (where the heat pulled from your cabin air is dumped). Not functionally any different than the electric compressor in your fridge at home doing the same thing. In this case, there is a mechanical load on the engine when the clutch for the AC engages to run the compressor, so your mileage will drop a bit since some of that horsepower is driving a compressor instead of the input to your transmission.In a broader sense, the AC uses gas in the sense that it has a circulating fluid that transitions back and forth between liquid and gas as part of the refrigeration cycle. In this context, it’s used, but not ‘used up’.BTW, that mini-heat exchanger for the cabin heater to tie to the engine coolant is why you’ll see advisory signs on long mountain uphill grades to not just switch off AC in hot weather, but turn the heater on full bore. It allows the cabin heater to act like a small auxiliary radiator and help keep the engine from overheating if it’s teetering on the edge due to some combination of age, water pump health, or vehicle load. Sweltering in the cabin then, but preferable to a boil-over.
Why does it cost more to heat households than to cool them?
People have already made quite a few good points, but one that is missing is the fundamental difference between cooling and heating. If you heat with electricity, you are converting electrical energy to heat energy. It takes a lot of electricity - meaning it’s expensive - to make heat.In contrast, when you are cooling an area, you aren’t ‘making’ anything, meaning that you aren’t ‘making’ cold. What you are doing instead is taking the excess heat from the living quarters and moving it outside. Next time you run your AC, put your hand near the outside part of the unit, and feel the heat coming out of it. That heat was in your house.It is far cheaper to move heat from one place to another than it is to make it.Incidentally, this is how refrigerators work - they move the heat from your food to your kitchen. In warm areas of the country, you can use a ‘heat pump’ which moves the heat from the outside of the house (even if it is 40 or 50F outside) to the inside. The colder it gets outside, the less well that works. Geothermal systems also work that way - since they don’t ‘make’ heat, but just collect it from the environment, they are cheap to operate.Oh, and if you’re ever thinking of buying one of those ‘stand alone air conditioners’ that don’t need to be vented, forget it. If they aren’t dumping the heat outside, you’re going to be very disappointed.As to whether or not the original assertion in the question is true, it varies enormously depending on the heat source that the home uses.
How much does it typically cost to fill a 20 lb propane tank in the US?
Fuel prices in America differ across the regions. As many factors are included. such as privatization of the Industry. Competitive market availability of the resources.Gallon is the standard weight term in United States to sale propane. 1 US liquid Gallon is equal to 3.79 litre and 1 litre is equal to 2.2 pound.According to today's rage average rate of propane today was $ 2.36 / Gallon Lets do simple maths here.20 Pound / 2.2 = 9.09 Litre9.09 Litre / 3.79 = 2.40Cost will be $ 2.36 x 2.40 Gallon = $ 5.664
A gas expands from 2 liters to 6 liters against a constant pressure of 0.5 ATM on absorbing 200J of heat. What is change in internal energy?
ΔH = ΔU + PΔV... (Definition of Enthalpy change)At constant pressure (for isobaric process)ΔH = q….Hence, ΔU = q - PΔV (1st law of thermodynamics)ΔU = 200 - 0.5*101325 *(0.006–0.002)ΔU = -2.65 JAll the sign conventions were considered and pressure and volume were converted to their respective SI units.1 atm pressure = 101325 Pa1 liter volume = 0.001 cubic m.
How much does utility bills cost in a one bedroom apartment?
1 million dollars. ok.. are you living alone? do you cook? do you love Air cond? Where do you live? single person water bill which usually comes with trash bill about $20-$25. Gas depends on if apartment uses Gas for stove and waterheater.. maybe $25 more in winter months Electricity depends on your electronic toys and how much you love AC.. could be $65-$250/month everthing depends on your lifestyle Cable $45-$100 Internet if not bundled with Cable $45 Cell phone $35-$100 Good luck.. some apartment require renters insurance.
An experiment to determine the specific heat of a gas makes use of a water manometer attached to a flask (the?
To calculate the difference between two different pressures, you use manometer. In this case it's the water manometer. The difference between these two pressures in your case is defined with the following equation: P_diff = ρ * g * z, where: * ρ is density of the material used in a manometer, * g is standard gravity, and * z is the difference between two levels of liquid In your case, if this difference is 1 cm (0.01 m), you will have: P_diff = ρ * g * z = 1000 * 9.81 * 0.01 = 98.1 Pa The unit is derived as: (kg/m^3) * (m/s^2) * m = kg/(m*s^2) = Pa Now I did not understand what "1.5 105 Pa" represents, but the difference between these two pressures is 98.1 Pa. If the left side of the manometer has lower liquid level than the right side, than the pressure on the left is higher than that on the right, and vice versa. Just add/subtract 98.1 to the pressure value you already know (depending on your case), and you'll get your result.
If 1.3 mol of an ideal gas absorbs 750 J heat while performing 625 J of work, what is its temp change?
You use equations for an ideal gas: pV = NkT = nRT U = CvNkT = Cv nRT where: Cv is a constant dependent on temperature (e.g. equal to 3/2 for a monatomic gas for moderate temperatures) U is the internal energy p is the pressure V is the volume n is the amount of gas (moles) R is the gas constant, 8.314 J·K^−1mol^-1 T is the absolute temperature N is the number of particles k is the Boltzmann constant, 1.381×10^−23 J·K^−1 In your first problem, the energy difference between heat absorbed by the gas and the work performed is the energy lost to heat. Thus, ∆E = Cv nR∆T where ∆T is the temperature change ∆T = ∆E / (Cv nR) = (750-625) / (3/2)(1.3)(8.314) = 7.71° Kelvin