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10-1.Thermometry, Thermal Expansion and Calorimetry
medium
Two liquids $A$ and $B$ are at $32\,^oC$ and $24\,^oC.$ When mixed in equal masses the temperature of the mixture is found to be $28\,^oC$. Their specific heats are in the ratio of
A
$3:2$
B
$2:3$
C
$1:1$
D
$4:3$
Solution
Heat lost by $A=$ heat gained by $\mathrm{B}$
$\Rightarrow m_{A} \times c_{A} \times\left(T_{A}-T\right)=m_{B} \times c_{B} \times\left(T-T_{B}\right)$
since $m_{A}=m_{B}$ and temperature of the mixture $(T)=28^{\circ} C$
$\therefore C_{A} \times(32-28)=c_{B} \times(28-24)$
$\Rightarrow \frac{c_{A}}{c_{B}}=1: 1$
Standard 11
Physics