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
\[
\begin{gather}
\bbox[#99CCFF,10px]
{Q=mc\Delta t} \tag{I}
\end{gather}
\]
\[
\begin{gather}
c=\frac{Q}{m(t_f-t_i)}\\[5pt]
c=\frac{2300\;\mathrm{cal}}{(250\;\mathrm g)(100-0)\;\mathrm{°C}}\\[5pt]
c=\frac{23\cancel{00}\;\mathrm{cal}}{250\cancel{00}\;\mathrm{g°C}}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{c=0.092\;\mathrm{cal/g°C}}
\end{gather}
\]
b) The heat capacity of the body is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{C=mc}
\end{gather}
\]
\[
\begin{gather}
C=(250\;\mathrm{\cancel g})\left(0.092\;\mathrm{\small{\frac{cal}{\cancel g°C}}}\right)
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{C=23\;\mathrm{cal/°C}}
\end{gather}
\]
c) If the body loses heat, we have Q = −1 000 cal, the temperature of 100 °C becomes the initial temperature of the body, ti, applying the equation (I)
\[
\begin{gather}
Q=mc(t_f-t_i)\\[5pt]
t_f-t_i=\frac{Q}{mc}\\[5pt]
t_f=\frac{Q}{mc}+t_i\\[5pt]
t_f=\frac{-1000\;\mathrm{\cancel{cal}}}{(250\;\mathrm{\cancel g})\left(0.092\;\mathrm{\frac{\cancel{cal}}{\cancel g°C}}\right)}+100\;\mathrm{°C}\\[5pt]
t_f=-\frac{1000}{23\;\mathrm{\frac{1}{°C}}}+100\;\mathrm{°C}\\[5pt]
t_f=(-43.5+100)\;\mathrm{°C}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{t_f=56.5\;\mathrm{°C}}
\end{gather}
\]
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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 .