Solved Problem on Heat
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A body made of 250 g of brass is heated from 0 °C to 100 °C, for this 2300 cal were used. Calculate:
a) The specific heat of the brass;
b) The heat capacity of the body;
c) If the body in the final situation loses 1000 cal, what will be its temperature?
Problem data:
- Mass of the body: m = 250 g;
- Initial temperature: ti = 0 °C;
- Final temperature: tf = 100 °C;
- Heat used to warm the body: Q = 2300 cal.
a) Using the equation for heat transfer, we can find the specific heat
\[ \bbox[#99CCFF,10px]
{Q=mc\Delta t}
\]
\[
\begin{gather}
c=\frac{Q}{m(t_{f}-t_{i})}\\
c=\frac{2300}{250\times (100-0)}\\
c=\frac{2300}{25000}\\
c=\frac{23}{250}
\end{gather}
\]
\[ \bbox[#FFCCCC,10px]
{c=0.092\;\text{cal/g°C}}
\]
b) The heat capacity of the body
\[ \bbox[#99CCFF,10px]
{C=mc}
\]
\[
C=250 \times 0.092
\]
\[ \bbox[#FFCCCC,10px]
{C=23\;\text{cal/}^{\text{o}}\text{C}}
\]
c) If the body loses heat we have Q = −1 000 cal, the temperature of 100°C becomes the initial temperature of the body and we want the final temperature
\[ \bbox[#99CCFF,10px]
{Q=mc\Delta t}
\]
\[
\begin{gather}
-1000=250 \times 0.092 \times (\;t_{f}-100\;)\\
-1000=23 \times (\;t_{f}-100\;)\\
t_{f}-100=\frac{-{1000}}{23}\\
t_{f}=-43.5+100
\end{gather}
\]
\[ \bbox[#FFCCCC,10px]
{t_{f}=56.5\;^{\text{o}}\text{C}}
\]
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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 .