Question

1. How much heat is necessary to vaporize 500 g of water at its boiling point?

Answer 1

1.    When water converts to vapor phase from liquid phase, there is no any temperature change. In other words, when a substance changes its phase, the temperature remains constant.

So we can use the formula

Q = mL

Where Q is the heat (J), m is the mass (g) and L is the latent heat (J/g).

 

M = 500 g

L = Latent heat of vaporization = 2230 J/g

 

Hence, Q = 500 g x 2230 J/g = 1115000 J = 1115 kJ

Hence, 1115 kJ of heat is needed to vaporize 500 g of water at its boiling point.

 

2.    Let’s assume that water is at its freezing point (0 ⁰C). Then there is only phase conversion and no any temperature change. Again we can use the formula,

 Q = mL

                      

Here, Q is the heat that given off (J), m is the mass (g) and L is the latent heat of fusion (J/g).

 

Q = 5100 J

L = 334 J/g

 

Hence, 5100 J = m x 334 J/g

                    m = 5100 J / 334 J/g

                    m = 15.269 g

Hence, the mass of water is 15.269 g.

3.  When condensing vapor into its liquid again it is only a phase conversion and temperature remains as constant (100 ⁰C).

Hence, we can use the formula,

 Q = mL

 

Here, Q is the heat that given off (J), m is the mass of vapor (g) and L is the Latent heat of vaporization (J/g).

 

Q = 57.500 J

L = 2230 J/g

By applying the equation,

          57.500 J = m x 2,230 J/g

                     m =0.0258 g

Hence, the mass of steam is 0.0258 g.