How Do Heating Costs and Battery Drain Calculations Work in Electrical Systems?

[Shadowsol]

1. Compute the cost per hour at 8 cents per kw-hr of electrically heating a room if it requires 1 kg per hour of anthracite coal having a heat of combustion of 8000 kcal/kg.

2. The lights on a car are left on. They dissipate 95 watts. How long will it take for the fully charged 12-volt battery to run down if the battery is rated a 150 ampere-hours (A-hr)?

3. What is the cost of electrically heating 50 liters of water(50kg) from 40 degrees C to 100 degrees C at 8 cents per kw-hr?







3. For the first 2 I honestly don't know how to start or what to do. For the 3rd one, I multiplied 50*4.18*60 to get the joules. Than I multiplied that by the 8 cents. The answer I got was 1 dollar. I'm not sure if this is right, the problem does not give the time it takes, so I don't know how the 8 cents per kw-hr would work.

 

[kamerling]

1. you know how much anthracite it takes per hour to heat the room. You know how much a kg of anthracite produces if you burn it. you can compute how much energy you need in an hour. you have to convert that to Kilowatthours. Mmake sure you get the units right.

1 kcal is the energy to heat 1 kg of water 1 degree celsius (or kelvin) and is equal to 4184 Joules.
1 kilowatr-hr is the energy that a powersource of 1 kilowatt produces in an hour.
it is equal to 3600000 Joules

2. compute how many amperes the lights use. In an hour they will use that amount of amperehours

3. 1 dollar is wrong.
- if you use kilograms of water you will get an answer in kilojoules. You must then determine how much Kilowatthours that energy is, and multiply that by the price of a kilowatthour.

 

[Moetasim]

1. To compute the cost per hour of electrically heating a room, we first need to calculate the energy required to heat the room. This can be done by converting the heat of combustion of anthracite coal from kcal/kg to kWh/kg. Using the conversion factor of 1 kcal = 0.001163 kWh, we get 8000 kcal/kg = 9.304 kWh/kg.

Since the room requires 1 kg per hour of anthracite coal, it would require 9.304 kWh of energy per hour. The cost per hour can then be calculated by multiplying the energy required (9.304 kWh) by the cost per kilowatt-hour (8 cents), which gives us 74.432 cents per hour.

2. To calculate how long it would take for the battery to run down, we need to use the formula P=VI, where P is power, V is voltage, and I is current. The power dissipated by the lights is given as 95 watts, and the voltage of the battery is 12 volts. Using the formula, we can calculate the current to be 7.92 amperes (A).

Since the battery is rated at 150 A-hr, we can determine that it would take 150/7.92 = 18.94 hours for the battery to run down.

3. To calculate the cost of electrically heating 50 liters (or 50 kg) of water, we first need to calculate the energy required to raise the temperature from 40 degrees C to 100 degrees C. This can be done using the formula Q=mCΔT, where Q is the energy, m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.

Plugging in the values, we get Q = (50 kg)(4.18 J/g°C)(60°C) = 12540 kJ. Converting this to kWh, we get 3.483 kWh.

Multiplying this by the cost per kilowatt-hour (8 cents), we get a total cost of 27.864 cents. However, this calculation assumes that the heating is done instantaneously. If the heating takes longer than an hour, the cost would be higher. Without knowing the time it takes to heat the water, we cannot accurately determine the cost.