Water (2410 gg ) is heated until it just begins to boil. If the water absorbs 4.95×105 JJ of heat in the process, what was the initial temperature of the water?

51.1°C

Explanation:

Parameters given:

Mass of water = 2410g = 2.41 kg

Heat absorbed = 4.95 * 10^5 J

The formula for heat capacity in terms of mass, specific heat capacity and temperature is given as:

H = mC(T2 - T1)

Where m = mass

C = specific heat capacity of water = 4200 J/kg°C

T2 = final temperature

T1 = initial temperature

In this case, the final temperature is 100°C since the water boils, hence, to find T1,

4.95 * 10^5 = 2.41 * 4290 * (100 - T1)

=> 100 - T1 = 495000/(2.41 * 4200)

100 - T1 = 48.9

=> T1 = 100 - 48.9

T1 = 51.1°C

The initial temperature of the water was 51.1°C

5 w for 1hr = 5x60x60 joules=300x60=3000x6=18,000j as total energy conversion. efficiency of bulb needed for radiant to thermal within 18000

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