Solved Problem on Heat
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A blender has a power of 90 W and is used to stir 200 g of water for 1 minute. The blender loses 10% of its energy through dissipation, the other 90% of the energy is the work transferred as heat to the water inside the blender. Calculate the heating of the mass of water. Given: 1 cal = 4.2 J.


Problem data:
  • Blender power:    P = 90 W;
  • Energy dissipated:    Ed = 10% E;
  • Mass of water:    m = 200 g;
  • Time interval:    Δt = 1 min;
  • Specific heat of water:    c = 1 cal/g°C;
  • Mechanical equivalent of heat:    1 cal = 4,2 J.
Problem diagram:

Of the total energy E produced by the electric motor, 10% is lost through dissipation Ed (such as heat and sound produced by the blender), and the other 90% of the energy is used in work W to produce heat Q which heats the water (Figure 1).
\[ \begin{gather} E-E_{d}=W=Q \end{gather} \]
Figure 1

Solution

Converting the time interval given in minutes (min) to seconds (s)
\[ \begin{gather} \Delta t=1\;\cancel{\text{min}}\times\frac{60\;\text{s}}{1\;\cancel{\text{min}}}=60\;\text{s} \end{gather} \]
The power delivered by the blender during the running time is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {P=\frac{E}{\Delta t}} \end{gather} \]
\[ \begin{gather} E=P\Delta t\\[5pt] E=90\times 60\\[5pt] E=5400\;\text{J} \end{gather} \]
The dissipated energy will be
\[ \begin{gather} E_{d}=10\text{%}E\\[5pt] E_{d}=\frac{10}{100}.5400\\[5pt] E_{d}=0.1\times 5400\\[5pt] E_{d}=540\;\text{J} \end{gather} \]
The energy used in the work to heat the water will be
\[ \begin{gather} W=E-E_{d}\\[5pt] W=5400-540\\[5pt] W=4860\;\text{J} \end{gather} \]
Converting the calculated energy in joules (J) to calories (cal) used in the problem
\[ \begin{gather} W=4860\;\cancel{\text{J}}\times\frac{1\;\text{cal}}{4.2\;\cancel{\text{J}}}=1157\;\text{cal} \end{gather} \]
\[ \begin{gather} W=Q=1157\;\text{cal} \end{gather} \]
The equation of heat is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {Q=mc\Delta\theta} \end{gather} \]

Note: The symbol θ was used for the temperature, so as not to confuse it with t used for the time interval in the power formula.

\[ \begin{gather} 1157=200\times 1\times\Delta \theta \\[5pt] \Delta \theta=\frac{1157}{200} \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {\Delta \theta \approx 5.8\;°\text{C}} \end{gather} \]
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .