Solved Problem on Electric Current
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Inside a container with 0.5 kg of water is placed a resistor with 2 Ω of electrical resistance conducting an electric current of 5 A for 7 min. Calculate the water temperature rise, assuming there is no change in state. Data: the specific heat of water is equal to 1 cal/g°C and mechanical equivalent of heat 1 cal = 4.2 J.


Problem data:
  • Mass of water:    m = 0.5 kg;
  • Resistor:    r = 2 Ω;
  • Electric current:    i = 5 A;
  • Heating time interval:    Δt = 7 mim;
  • Specific heat of water:    c = 1 cal/g°C;
  • Mechanical equivalent of heat:    1 cal = 4.2 J.
Problem diagram:

When the electric current i flows through the circuit, the resistor produces a heat Q, by Joule heating, which will heat the water, making the temperature rise from a value Δθ (Figure 1).

Note: We use the symbol θ for the temperature, not to be confused with t used for time.
Figure 1

Solution

First, we must convert the interval of time given in minutes (min) for seconds (s) used in the International System of Units (S.I.).
\[ \Delta t=7\;\cancel{\text{min}}\times \frac{60\;\text{s}}{1\;\cancel{\text{min}}}=420\;\text{s} \]
The power dissipated in the resistor is given by
\[ \bbox[#99CCFF,10px] {\mathscr{P}=ri^{2}} \]
\[ \begin{gather} \mathscr{P}=2\times 5^{2}\\ \mathscr{P}=2\times 25\\ \mathscr{P}=50\;\text{W} \end{gather} \]
Power is the amount of energy transferred per unit of time, the energy dissipated in the resistor will be
\[ \bbox[#99CCFF,10px] {\mathscr{P}=\frac{E}{\Delta t}} \]
\[ \begin{gather} E=\mathscr{P}\Delta t\\ E=50\times 420\\ E=21000\;\text{J} \end{gather} \]
Converting the energy given in Joules (J), used in the International System of Units (S.I.), to calories (cal) and the water mass given in kilograms (kg) to grams (g), we have
\[ \begin{gather} E=Q=21000\;\cancel{\text{J}}\times \;\frac{1\;\text{cal}}{4.2\;\cancel{\text{J}}}=5000\;\text{cal}\\[10pt] m=0,5\;\cancel{\text{kg}}\times \;\frac{1000\;\text{g}}{1\;\cancel{\text{kg}}}=500\;\text{g} \end{gather} \]
Note: We convert the energy from Joules to calories because it is a more convenient unit for this problem. If the energy was left in Joules, we would have to convert the specific heat from the water, given in calories per gram degree Celsius (cal/g °C), to Joules per kilogram degree Celsius (J/kg °C). But usually, in the problems involving heat, which is energy in transit, it is given in calories.

The heat received by the water, where there is no change of state, will be
\[ \bbox[#99CCFF,10px] {Q=mc\Delta \theta} \]
the change of water temperature will be
\[ \begin{gather} \Delta \theta =\frac{Q}{mc}\\ \Delta \theta=\frac{5000}{500\times 1}\\ \Delta \theta =\frac{5000}{500} \end{gather} \]
\[ \bbox[#FFCCCC,10px] {\Delta \theta =10 °\text{C}} \]
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