A container at $0^{\circ}$ temperature contains $300 \mathrm{~g}$ of water and $50 \mathrm{~g}$ of ice; the container weighs $100 \mathrm{~g}$, and the specific heat of the material of the container is 0.5. How many grams of $100^{\circ}$ steam must be introduced into this container to make the final temperature $20^{\circ}$?
It is necessary that the heat loss of the steam be equal to the heat gain of the other materials. The steam first loses its latent heat of vaporization; in addition, the heat corresponding to $100-20=80$ degrees. On the other hand, the water and the container gain the heat corresponding to 20 degrees, the ice melts and gains the heat corresponding to 20 degrees; and thus

$$
536 x+80 x=300 \cdot 20+50 \cdot 80+50 \cdot 20+100 \cdot 0.5 \cdot 20
$$

and from this

$$
x=19.48 \mathrm{gr} .
$$

(Riesz Frigyes, Gymn. VIII. o. t., Györ.)