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:    \( \mathscr P = 90 W \);
  • 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)
\[ \begin{gather} \Delta t=1\;\mathrm{\cancel{min}}\times\frac{60\;\mathrm s}{1\;\mathrm{\cancel{min}}}=60\;\mathrm s \end{gather} \]
The power delivered by the blender during the running time is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {\mathscr P=\frac{E}{\Delta t}} \end{gather} \]
\[ \begin{gather} E=\mathscr P\Delta t \\[5pt] E=(90\;\mathrm W)(60\;\mathrm s)\\[5pt] E=5400\;\mathrm J \end{gather} \]
The dissipated energy will be
\[ \begin{gather} E_d=10\text{%}E \\[5pt] E_d=\frac{10}{100}\times 5400\;\mathrm J \\[5pt] E_d=0.1\times 5400\;\mathrm J \\[5pt] E_d=540\;\mathrm J \end{gather} \]
The energy used in the work to heat the water will be
\[ \begin{gather} W=E-E_d \\[5pt] W=(5400-540)\;\mathrm J \\[5pt] W=4860\;\mathrm J \end{gather} \]
Converting the calculated energy in joules (J) to calories (cal) used in the problem
\[ \begin{gather} W=4860\;\mathrm{\cancel J}\times\frac{1\;\mathrm{cal}}{4.2\;\mathrm{\cancel J}}\approx 1157\;\mathrm{cal} \end{gather} \]
\[ \begin{gather} W=Q=1157\;\mathrm{cal} \end{gather} \]
The equation of heat is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {Q=mc\Delta\theta} \end{gather} \]
\[ \begin{gather} \Delta\theta=\frac{Q}{mc} \\[5pt] \Delta\theta=\frac{1157\;\mathrm{\cancel{cal}}}{(200\;\mathrm{\cancel g})\left(1\;\mathrm{\frac{\cancel{cal}}{\cancel g°C}}\right)} \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {\Delta \theta \approx 5.8\;\mathrm{°C}} \end{gather} \]

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.
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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 .