Answer :
Final answer:
The energy required to heat 190kg of water from 15°C to 85°C can be calculated using the specific heat of water and the temperature difference. The energy is converted to kilowatt-hours, then multiplied by the cost per kWh, resulting in a total cost of approximately $0.93.
Explanation:
To answer this question, we need to calculate the energy required to heat the water, then convert that to cost. Specific heat of water is 4.184 J/g°C, which is the energy required to raise the temperature of 1 gram of water by 1°C. Since we're working with kg, note that 1 kg = 1000 g. Also, the temperature difference is 85°C - 15°C = 70°C.
Using the formula for energy, Energy = mass x specific heat x change in temperature, and plugging in relevant values, we get Energy = 190,000 g x 4.184 J/g°C x 70°C = 5.59 x 10^7 J.
The next step involves converting Joules to kilowatt-hours since we're given the energy cost in kilowatt-hours. Note that 1 kilowatt-hour (kWh) equals 3.6 x 10^6 Joules. So, the energy required in kWh is 5.59 x 10^7 J/3.6 x 10^6 J/kWh = 15.52 kWh.
Finally, to calculate the cost, multiply the energy in kWh by the cost per kWh: 15.52 kWh x $0.06/kWh = $0.93. So, it would cost approximately $0.93 to heat 190 kg of water from 15°C to 85°C.
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