Physics - Heat & Thermo Question with Solution | TestHub

When $A\&B$ are mixed

${\mathrm{m}}_1{ \mathrm{S}}_1(16-10)={\mathrm{m}}_2{ \mathrm{S}}_2(20-16)$

$6{ \mathrm{m}}_1{ \mathrm{S}}_1=4{ \mathrm{m}}_2{ \mathrm{S}}_2 .......\left(1\right)$

When $\mathrm{B}\&\mathrm{C}$ are mixed

${\mathrm{m}}_2{ \mathrm{S}}_2(26-20)={\mathrm{m}}_3{ \mathrm{S}}_3(30-26)$

$6{ \mathrm{m}}_2{ \mathrm{S}}_2=4{ \mathrm{m}}_3{ \mathrm{S}}_3 ...........\left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$

$9{ \mathrm{m}}_1{ \mathrm{S}}_1=4{ \mathrm{m}}_3{ \mathrm{S}}_3$

When $\mathrm{A}\&\mathrm{C}$ are mixed

${\mathrm{m}}_1{ \mathrm{S}}_1( \mathrm{T}-10)={\mathrm{m}}_3{ \mathrm{S}}_3(30-\mathrm{T})$

$\mathrm{T}=\frac{30{\mathrm{m}}_3{\mathrm{S}}_3+10{\mathrm{m}}_1{\mathrm{S}}_1}{{\mathrm{m}}_1{\mathrm{S}}_1+{\mathrm{m}}_3{\mathrm{S}}_3}$

$\mathrm{T}=\frac{10\left(3{\mathrm{m}}_3{\mathrm{S}}_3+{\mathrm{m}}_1{\mathrm{S}}_1\right)}{{\mathrm{m}}_3{\mathrm{S}}_3+{\mathrm{m}}_1{\mathrm{S}}_1}$

$\mathrm{T}=\frac{10\left(3{\mathrm{m}}_3{ \mathrm{S}}_3+\frac{4}{9}{ \mathrm{m}}_3{ \mathrm{S}}_3\right)}{{\mathrm{m}}_3{ \mathrm{S}}_3+\frac{4}{9}{ \mathrm{m}}_3{ \mathrm{S}}_3}$

$\mathrm{T}=\frac{10\times 31}{13}$

$=23.84°\mathrm{C}$