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

666,480 Joules or 669.48 kJ

Explanation:

We are given;

but, density of water is 1 g/ml

Required 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

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