The mass of water in the kettle is 0.50 kg. The specific heat capacity of water is 4200 J/kg °C. The initial temperature of the water is 100 °C. Calculate the average power output from the water in the kettle to the surroundings in 2 hours

1.27 Mega watt approx

Explanation:

Given data

Mass= 0.5kg

c= 4200 J/kg °C

T1= 100°C

Time = 2 hours >>>> 7200 seconds

From the given data the water is already at the boiling point

Hence if continuous heat is being added the latent heat of vapourization will result, the latent heat of vaporization is about 2,260 kJ/kg

Q= mL

Q= 0.5*2,260

Q = 1130kJ

Power= Energy * Time

Power=  1130* 1130

Power= 1.27mega watt approx

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west

causes

fossil fuels burn

cities become more industrialized

human population grows worldwide