Q = mc∆T
Where:
Q is the heat energy applied (6x10^15 joules)
m is the mass of water (in kilograms)
c is the specific heat capacity of water (4186 J/kg°C)
∆T is the change in temperature (1 degree Celsius)
Solving for m:
Q = mc∆T
m = Q / (c∆T)
m = (6x10^15 J)/(4186 J/kg°C X 1°C)
m = 1.43x10^12 kg
Now, we need to convert the mass of water to volume. The density of water at room temperature is approximately 1000 kg/m^3.
So, the volume of water that can be heated by 1 degree when 6x10^15 joules are applied is:
V = m/ρ
V = (1.43x10^12 kg)/(1000 kg/m^3)
V = 1.43x10^9 m^3
Therefore, 6x10^15 joules of heat energy can heat 1.43x10^9 cubic meters of water by 1 degree Celsius.
