The same current flows through a copper voltameter and a wire with a resistance of $10 \mathrm{ohm}$. In the voltameter, 1.1081 grams of copper are deposited on the negative electrode in 30 minutes. Calculate how many gram-calories of heat are developed in the wire. The electrochemical equivalent of copper is 0.0003275 (ampere minute).
The electrochemical equivalent means that a 1 A current extracts how many grams of the respective material in 1 second. Therefore, the number of amperes that have passed is obtained by dividing the amount of substance extracted in one second by the electrochemical equivalent.

$$
I=\frac{1.1081}{30 \cdot 60 \cdot 0.0003275} \mathrm{~A}=1.88 \mathrm{~A}
$$

Therefore, the amount of heat generated per second is

$$
H=I^{2} \cdot r \cdot\left\{\mathrm{A}^{2} \cdot \Omega\right\}
$$

since

$$
1 \mathrm{~A}=\frac{1 \cdot \mathrm{V}}{1 \cdot \Omega}
$$

thus

$$
1 \Omega=\frac{1 \cdot \mathrm{V}}{1 \mathrm{~A}}
$$

and so

$$
\begin{gathered}
H=I^{2} \cdot r\left\{A^{2} \cdot \frac{V}{A}\right\}=I^{2} \cdot r \text { voltampère }=I^{2} r \text { watt } \\
1 \text { watt }=0.24 \frac{\text { gramcalorie }}{\mathrm{sec}} . \\
H=I^{2} \cdot r \cdot 0.24(\mathrm{grcal})=1.88^{2} \cdot 10 \cdot 0.24(\mathrm{grcal}) \\
H=8.48(\mathrm{grcal}) .
\end{gathered}
$$

(Richárd Kürth, Nyitra.)