A room contains 47kg of air. How many kilowatt-hours of energy are necessary to heat the air in the house from 7∘C to 26∘C? The heat capacity of air is 1.03 J/g∘C
q = mass air x specific heat air x (Tfinal-Tintial)
You know q, mass, Tf and Ti. You will need to look up the specific heat of air. Solve for q (in joules) needed.
Then 1 watt = 1 joule/second
Here is a site that will convert joules to kW-hr.
http://www.rapidtables.com/convert/energy/Joule_to_kWh.htm
0.0012
To calculate the amount of energy required to heat the air in the room, we can use the following steps:
Step 1: Find the mass of the air in the room in grams
Given that the mass of air in the room is 47 kg, we need to convert it to grams.
1 kg = 1000 grams
Therefore, 47 kg = 47,000 grams.
Step 2: Calculate the temperature difference
The temperature difference is the final temperature minus the initial temperature. In this case, the final temperature is 26°C and the initial temperature is 7°C.
Temperature difference = 26 - 7 = 19°C.
Step 3: Convert the heat capacity to joules
The heat capacity of air is given as 1.03 J/g∘C. Since we have the mass in grams, no conversion is necessary.
Step 4: Calculate the total energy required
To find the total energy required, we multiply the mass, temperature difference, and heat capacity together.
Total energy = mass × temperature difference × heat capacity
Total energy = 47,000 g × 19°C × 1.03 J/g∘C
Calculating the total energy:
Total energy = 958,510 J
Step 5: Convert joules to kilowatt-hours
Since 1 joule is equal to 0.0002778 kilowatt-hours,
Total energy (in kilowatt-hours) = 958,510 J × 0.0002778 kWh/J
Total energy (in kilowatt-hours) ≈ 0.266 kWh
Therefore, approximately 0.266 kilowatt-hours of energy are necessary to heat the air in the room from 7°C to 26°C.