Gujarati
Hindi
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

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