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How many kilojoules of energy are required to bring 800 g of water from 20 degrees Celsius to the boiling point?

User Motox
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Final answer:

The specific heat of water is 4.184 J/g°C. To heat 800 g of water from 21°C to 85°C, we can use the equation: Energy = specific heat capacity x mass x temperature change. The energy required is 210 kJ.

Step-by-step explanation:

The specific heat of water is 4.184 J/g °C, so to heat 1 g of water by 1 °C requires 4.184 J. We note that since 4.184 J is required to heat 1 g of water by 1 °C, we will need 800 times as much to heat 800 g of water by 1 °C. Finally, we observe that since 4.184 J are required to heat 1 g of water by 1 °C, we will need 64 times as much to heat it by 64 °C (that is, from 21 °C to 85 °C).

This can be summarized using the equation:

Energy = specific heat capacity x mass x temperature change

Energy = 4.184 J/g °C x 800 g x 64 °C = 210,000 J (= 210 kJ)

Because the temperature increased, the water absorbed heat and q is positive.

User Morganbaz
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