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 2.99 kg of steam at 100 °C releases when it condenses?

Answer 1

Question #70930

The latent heat of fusion for water is $33.5 \times 10^{4} \, \mathrm{J/kg}$, while the latent heat of vaporization is $22.6 \times 10^{5} \, \mathrm{J/kg}$. What mass m of water at $0{}^{\circ} \mathrm{C}$ must be frozen in order to release the amount of heat that $2.99 \, \mathrm{kg}$ of steam at $100{}^{\circ} \mathrm{C}$ releases when it condenses?

Solution

As the amounts of heat released by condensation and fusion are equal we may say that:

$$\mathrm{Lm}_1 = \lambda \mathrm{m}_2, \text{ where}$$

- L -- the latent heat of vaporization for water;

- $\lambda$ -- the latent heat of fusion for water;

- $\mathrm{m}_1$ -- mass of water condensed;

- $\mathrm{m}_2$ -- mass of water must be frozen.

$$\mathrm{m}_2 = \frac{\mathrm{Lm}_1}{\lambda}$$ $$\mathrm{m}_2 = \frac{2.99 \times 22.6 \times 10^5}{33.5 \times 10^4} = 20.17 \, (\mathrm{kg})$$

Answer

20.17 kg of water at $0{}^{\circ} \mathrm{C}$ must be frozen in order to release the amount of heat that 2.99 kg of steam at 100 °C releases when it condenses

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