Solved Problem on Heat

A blender has a power of 84 W and is used to stir 100 g of water for 30 seconds. It is assumed that all the work is transferred as heat to the water inside the blender and that there are no other energy transfers. Calculate the heating of the mass of water. Given: 1 cal = 4.2 J.

Problem data:

Blender power: \( {\mathscr P} = 84 W \);
Mass of water: m = 100 g;
Time interval: Δt = 30 s;
Specific heat of water: c = 1 cal/g°C;
Mechanical equivalent of heat: 1 cal = 4.2 J.

Problem diagram:

All the energy E produced by the electric motor is used in work W to produce heat Q, which heats the water (Figure 1).

\[ \begin{gather} E=W=Q \end{gather} \]

Figure 1

Solution

The power delivered by the blender during the running time is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {{\mathscr P}=\frac{E}{\Delta t}} \end{gather} \]

\[ \begin{gather} E=\mathscr{P}\Delta t\\[5pt] E=84\times 30\\[5pt] E=2520\;\mathrm{J} \end{gather} \]

Converting the calculated energy in joules (J) to calories (cal) used in the problem

\[ \begin{gather} E=2520\;\cancel{\mathrm{J}}\times\frac{1\;\mathrm{cal}}{4.2\;\cancel{\mathrm{J}}}=600\;\mathrm{cal} \end{gather} \]

\[ \begin{gather} E=Q=600\;\mathrm{cal} \end{gather} \]

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.

\[ \begin{gather} 600=100\times 1\times \Delta \theta \\[5pt] \Delta \theta=\frac{600}{100} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {\Delta \theta =6\;°\mathrm{C}} \end{gather} \]

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