How much energy is need to go heat 20.0 grams of water from 25.0 degrees celsius to 50.0 degrees celsius

Okay, we first need to calculate the amount of heat energy required to raise the temperature of water to its boiling point(100°C) using the formula Q= mcθ where Q is the amount of heat energy required, m is the mass of the sample of water, c is the specific heat capacity of water and θ is the change in temperature.

Q= (20.0)(0.999043976)(100-25)

= 1.498565965x10³ cal

We now need to calculate the amount of heat energy required to convert the sample of water to steam at its boiling point using the formula Q= mL where Q is the amount of heat energy required, m is the mass of the sample of water and L is the latent heat of vapourisation of water.

Q= (20.0)(540)

= 1.08x10⁴ cal

Now, we shall calculate the amount of heat energy required to raise the temperature of the steam to 150°C using the formula Q= mcθ.

Q= (20.0)(0.485)(150-100)

= 485 cal

Finally, we add up the amount of heat energy required at each of the three stages to determine the net amount of heat energy required, Qₙ.

Qₙ= (1.498565965x10³)+(1.08x10⁴)+(485)

= 1.278356597x10⁴ cal

= 1.28x10⁴ cal correct to 3 significant figures.

the objects have the same charge.