an immersion heat can supply heat at a rate of 5.2 x 10 to the power of 2 joules per second. calculate the time that it wil take to completly evaporate 1.25 x 10 to the power of minus 1 kg water initially at a temperature of 21 degrees celsius
Power*time= Heat= Mass*Hvwater + Mass*(100-21)*specific heat of water.
Watch specific heat, heat of vaporization units, make certain they match your mass units.
Solve for time.
To calculate the time it will take to completely evaporate the water, we need to use the equation:
Q = mL
Where:
Q is the amount of heat required to evaporate the water,
m is the mass of the water, and
L is the latent heat of vaporization of water (which is approximately 2.26 x 10^6 J/kg).
First, let's convert the initial mass of water from kilograms to grams:
Mass = 1.25 x 10^-1 kg = 1.25 x 10^-1 kg * 1000 g/kg = 1.25 x 10^2 g
Next, let's calculate the amount of heat required to evaporate the water:
Q = mL
Q = (1.25 x 10^2 g) * (2.26 x 10^6 J/kg)
Q = 2.825 x 10^8 J
Now, let's calculate the time it will take for the immersion heater to supply this amount of heat:
Power = 5.2 x 10^2 J/s
Time = Q / Power
Time = (2.825 x 10^8 J) / (5.2 x 10^2 J/s)
Time = 5.44 x 10^5 s
The time it will take to completely evaporate the water is approximately 5.44 x 10^5 seconds.