A pot containing 2 liters of 5 celsius water is putt on a stove that is 1500 watts. Now the stove is on. How long time will it take for all of the water to boil from the pot? (turn into vapour). No heat is lost to the surroundings.
a liter of water is a kilogram
1500 t = Joules in if t is seconds
1500 t = specific heat water * 2 kg *(100-5)degrees K + heat of vaporization* 2 kg
To calculate the time it takes for the water to boil, you will need to consider the amount of heat required to raise the water temperature from 5 degrees Celsius to its boiling point and the power output of the stove.
The heat required to raise the temperature of the water can be calculated using the specific heat capacity of water:
Q = mcΔT,
where Q is the heat energy, m is the mass of water in kilograms, c is the specific heat capacity of water (which is approximately 4.18 kJ/kg°C), and ΔT is the change in temperature.
First, you need to convert the volume of water (2 liters) to its mass. The density of water is approximately 1 gram/mL, so 2 liters of water would weigh 2000 grams or 2 kilograms.
Next, calculate the heat required to raise the temperature from 5°C to the boiling point of water, which is 100°C:
Q = 2 kg × 4.18 kJ/kg°C × (100°C - 5°C).
Once you have calculated the total heat required, you can determine the time it takes for the water to boil using the power output of the stove:
Time = Q / Power.
The power output of the stove is given as 1500 watts.
Let's perform the calculations:
Q = 2 kg × 4.18 kJ/kg°C × (100°C - 5°C) = 790 kJ.
Time = 790 kJ / 1500 watts.
Converting the kilowatts to seconds:
Time = 790 kJ / (1500 W × 1000 J/kJ) = 0.527 seconds.
Therefore, it will take approximately 0.527 seconds for all of the water to boil from the pot, assuming no heat is lost to the surroundings.