Solved Problem on Heat
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A body of mass 200 g is heated for 30 seconds by an energy source that provides a power of 210 W at a constant rate. Given the graph of temperature as a function of time, determine the heat capacity of the body, knowing that 1 cal = 4.2 J.


Problem data:
  • Mass of the body:    m = 200 g;
  • Power of the heat source:   \( \mathscr{P} \) = 210 W.
Solution:

First, we must convert the power unit given in watts (joules per second) to calories per second (cal/s), in this problem, it is more convenient not to use the International System of Units (SI)
\[ \begin{gather} \mathscr P=210\;\mathrm W=210\;\mathrm{\frac{J}{s}}=210\;\mathrm{\cancel J}\;\frac{1\;\mathrm{cal}}{4,2\;\mathrm{\cancel J}}\times\frac{1}{\mathrm s}=50\;\mathrm{\frac{cal}{s}} \end{gather} \]
The heat capacity is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {C=mc} \tag{I} \end{gather} \]
The quantity of heat received (or lost) is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {Q=mc\Delta \theta } \tag{II} \end{gather} \]
where θ was used for temperature in place of t was used for the time in the problem. Substituting the expression (I) into (II)
\[ \begin{gather} Q=C \Delta\theta \\[5pt] C=\frac{Q}{\Delta\theta} \tag{III} \end{gather} \]
During the heating time, by the graph, we see that the temperature varied from
\[ \begin{gather} \Delta\theta=\theta_f-\theta_i \\[5pt] \Delta\theta=(25-15)\;\mathrm{°C} \\[5pt] \Delta\theta=10\;\mathrm{°C} \tag{IV} \end{gather} \]
The heat received by the body is obtained from the power of the energy source. The source provides 50 calories in 1 second, so the 30 second heating will provide a quantity Q of heat, using Cross-multiplication
\[ \begin{gather} \frac{50\;\mathrm{cal}}{1\;\mathrm s}=\frac{Q}{30\;\mathrm s} \end{gather} \]
performing the cross multiplication
\[ \begin{gather} (1\;\mathrm s)\times(Q)=(30\;\mathrm s)\times(50\;\mathrm{cal}) \\[5pt] Q=\frac{(30\;\mathrm{\cancel s})\times(50\;\mathrm{cal})}{1\;\mathrm{\cancel s}} \\[5pt] Q=1500\;\mathrm{cal} \tag{V} \end{gather} \]
substituting the values (IV) and (V) into equation (III), the heat capacity will be
\[ \begin{gather} C=\frac{150\cancel 0\;\mathrm{cal}}{1\cancel 0\;\mathrm{°C}} \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {C=150\;\mathrm{cal/°C}} \end{gather} \]
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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 .