Question

The latent heat of fusion for water is 33.5 × 10⁴ J/kg, while the latent heat of vaporization is 22.6 × 10⁵ J/kg. What mass m of water at 0 °C must be frozen in order to release the amount of heat that 2.48 kg of steam at 100 °C releases when it condenses?

Answer 1

To find the mass m of water that must be frozen to release the same amount of heat as 2.48 kg of steam condensing, we can use the equation Qv = mLv, where Qv is the heat released by condensation and Lv is the latent heat of vaporization. Solving for m, we find that 0.0148 kg of water at 0 °C must be frozen.

Step-by-step explanation:

To find the mass m of water that must be frozen in order to release the same amount of heat as 2.48 kg of steam condensing, we can set up an equation using the latent heat of fusion and vaporization:

Qv = mLv

Where Qv is the heat released by condensation and Lv is the latent heat of vaporization. The heat released by condensing 2.48 kg of steam is given as:

Qv = m(2256 kJ/kg)

We can solve for m by rearranging the equation:

m = Qv / Lv

Substituting in the given values:

m = (2.48 kg)(2256 kJ/kg) / (33.5 × 10⁴ J/kg)

Simplifying the calculation:

m = 0.0148 kg

Therefore, 0.0148 kg of water at 0 °C must be frozen to release the same amount of heat as 2.48 kg of steam condensing.

Answer 2

16.78kg

Step-by-step explanation:

Data Given from question

Heat of fusion of water Lf=33.5×10^4 J/kg Lf=33.5×10^4 J/kg

Heat of vaporization Lv=22.6×10^5 J/kg Lv=22.6×10^5 J/kg

Mass of the steam condensed Ms=2.48 kgms=2.48 kg

From the condition given in the question

Heat Released during freezing = Heat released during condensation

Where Mw= Mass of water

MwLf=MsLv

MwLf=MsLv

Mw=MsLv÷Lf

Mw=MsLv÷Lf

Mw=2.48 kg×22.6×10^5 J/KG÷ 33.4×10^4 J/kg

Mw=2.48 kg×22.6×10^5 J/kg ÷33.4×10^4 J/kg

Therefore mass of water would be

Mw=16.78 kg