Answered

Sumali sa IDNStudy.com at tuklasin ang komunidad ng pagbabahagi ng kaalaman. Ang aming mga eksperto ay handang magbigay ng malalim na sagot at praktikal na solusyon sa lahat ng iyong mga tanong.

Electrical power companies sell electrical energy by kilowatt where 1 KW = 3.6 x 10€ J. Suppose that it cost Php 11.32 per kWh to run electric water heater. How much does it cost to heat 8.5 kg of water
from 18°C to 43°C to fill a bath tub. (Specific heat capacity of water is 4180 J/kg K)

Sagot :

Martusjenylynidn

Answer:

To solve this problem, we need to calculate the energy required to heat the water and then convert it to cost in PHP.

1. Energy Required to Heat Water:

  • Mass of water: 8.5 kg
  • Initial temperature: 18°C
  • Final temperature: 43°C
  • Specific heat capacity of water: 4180 J/kg K

Energy required to heat water = mass of water * specific heat capacity * temperature change

= 8.5 kg * 4180 J/kg K * (43°C - 18°C)

= 8.5 kg * 4180 J/kg K * 25 K

= 88,550 J

Convert energy from Joules to kilowatt-hours (kWh):

  • 1 kWh = 3,600,000 J
  • Energy required to heat water (in J) / 3,600,000 J/kWh

= 88,550 J / 3,600,000 J/kWh

≈ 0.0245 kWh

2. Cost to Heat Water:

  • Cost per kWh: PHP 11.32
  • Energy required to heat water (in kWh): 0.0245 kWh
  • Cost to heat water = cost per kWh * energy required to heat water

= PHP 11.32/kWh * 0.0245 kWh

≈ PHP 0.278

Therefore, it costs approximately PHP 0.278 to heat 8.5 kg of water from 18°C to 43°C to fill a bath tub.