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he0gaubong
17.03.2020 •
Chemistry
Calculate the heat change involved when 2.00 L of water is heated from 20.0/C to 99.7/C in
an electric kettle. (667 kJ)
I need the answer asap. Please help
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Ответ:
666,480 Joules or 669.48 kJ
Explanation:
We are given;
Volume of water as 2.0L or 2000 mlbut, density of water is 1 g/ml
Therefore, mass of water is 2000 g Initial temperature as 20 °C Final temperature as 99.7° CRequired to determine the heat change
We know that ;
Heat change = Mass × Temperature change × specific heat
In this case;
Specific heat of water is 4.2 J/g°C
Temperature change is 79.7 °C
Therefore;
Heat change = 2000 g × 79.7 °C × 4.2 J/g°C
= 669,480 Joules 0r 669.48 kJ
Thus, the heat change involved is 666,480 Joules or 669.48 kJ
Ответ: