Find the amount of heat needed to warm up clean H2O from 20 degrees celsius to vapor of 115 degrees celsius Ch20 = 4200 J/KG

To calculate the amount of heat needed to warm up water from 20 degrees Celsius to vapor of 115 degrees Celsius, we need to consider two steps: heating the water from 20 degrees to 100 degrees (the boiling point of water), and then vaporizing the water at 100 degrees to steam at 115 degrees.

1. First, calculate the heat needed to raise the temperature of water from 20 degrees to 100 degrees. The formula for this is Q = mcΔT, where Q is the heat energy, m is the mass of the substance (in this case, water), c is the specific heat capacity of the substance, and ΔT is the change in temperature.

Assuming we have 1 kg of water, the calculation would be:

Q1 = 1 kg * 4200 J/kg°C * (100°C - 20°C) = 336,000 J

2. Next, calculate the heat needed to vaporize the water at 100 degrees to steam at 115 degrees. The heat of vaporization of water is approximately 2260 kJ/kg.

Q2 = 1 kg * 2260,000 J/kg = 2260,000 J

3. Finally, add the two amounts of heat together to get the total heat needed:

Q_total = Q1 + Q2 = 336,000 J + 2260,000 J = 2596,000 J

So, the amount of heat needed to warm up 1 kg of clean water from 20 degrees Celsius to vapor of 115 degrees Celsius is approximately 2596,000 Joules.