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10-1.Thermometry, Thermal Expansion and Calorimetry
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$0.1\,m^3$ of water at $80\,^oC$ is mixed with $0.3\,m^3$ of water at $60\,^oC$. The final temperature of the mixture is ........ $^oC$
A
$65$
B
$70$
C
$60$
D
$75$
Solution
Let $T^{\circ} C$ be final temperature of the mixture.
According to principle of calorimetry,
Heat lost $=$ Heat gained
$\therefore \quad 0.1 \times 10^{3} \times s_{\text {water}} \times(80-T)$
$=0.3 \times 10^{3} \times s_{\text {water}} \times(T-60)$
$\Rightarrow T=65^{\circ} \mathrm{C}$
Standard 11
Physics