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

669.48 kJ

Explanation:

According to the question, we are required to determine the heat change involved.

We know that, heat change is given by the formula;

Heat change = Mass × change in temperature × Specific heat

In this case;

Change in temperature = Final temp - initial temp

                                       = 99.7°C - 20°C

                                       = 79.7° C

Mass of water is 2000 g ( 2000 mL × 1 g/mL)

Specific heat of water is 4.2 J/g°C

Therefore;

Heat change = 2000 g × 79.7 °C × 4.2 J/g°C

                      = 669,480 joules

But, 1 kJ = 1000 J

Therefore, heat change is 669.48 kJ

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