### Solved Problem on Heat

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A body made of 250 g of brass is heated from 0°C to 100°C, it is used 2300 cal to heat the body. Find:
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:    $$m=250\;\text{g}$$;
• initial temperature:    $$t_{\text{i}}=0\;^{\text{o}}\text{C}$$;
• final temperature:    $$t_{\text{f}}=100\;^{\text{o}}\text{C}$$;
• heat used to warm the body:    $$Q=2300\;\text{cal}$$.
Solution

a) Using the relationship between heat and temperature change we can find the specific heat
$\bbox[#99CCFF,10px] {Q=mc\Delta t}$
$c=\frac{Q}{m(\;t_{\text{f}}-t_{\text{i}}\;)}\\ c=\frac{2300}{250(\;100-0\;)}\\ c=\frac{23 \cancel{{00}}}{250 \cancel{{00}}}\\ c=\frac{23}{250}$
$\bbox[#FFCCCC,10px] {c=0.092\;\text{cal/g}^{\text{o}}\text{C}}$

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
$\bbox[#99CCFF,10px] {Q=mc\Delta t}$
$-1000=250 \times 0.092 \times (\;t_{\text{f}}-100\;)\\ -1000=23 \times (\;t_{\text{f}}-100\;)\\ t_{\text{f}}-100=\frac{-{1000}}{23}\\ t_{\text{f}}=-43.5+100$
$\bbox[#FFCCCC,10px] {t_{\text{f}}=56.5\;^{\text{o}}\text{C}}$