Answered

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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.

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