What is the mass of liquid water required to absorb #4.32*10^5# #kJ# of heat energy upon boiling?

I've been recently studying this in chemistry so I'm giving this a punt.

We can use the following formula

E=mcΔT

We assume that the (c) specific heat capacity of water is 4.184

The standard room temperature of water is 20 degrees, the boiling temperature is 100 degrees this results in a ΔT (change in temperature) of 80 degrees.

Finally we are given the amount of heat, 4.3210^5 kJ, this needs to be converted to Joules as that is the standard measurement this results in 4.3210^8 J

Can now solve for unknown m

4.32*10^8 = m4.184(80)

Then a quick division gives us the answer of m = 1290630 or 1290kg

The mass of liquid water required to absorb 4.32 x 10^5 kJ of heat energy upon boiling depends on the specific heat capacity of water and the heat of vaporization of water. The heat of vaporization of water is approximately 2260 kJ/kg. Therefore, you can use the formula:

Mass = (Heat energy) / (Heat of vaporization)

Mass = (4.32 x 10^5 kJ) / (2260 kJ/kg)