5 kg of ${\text{H}}_{\text{2}}\text{O}$  at $\text{10°C}$  is added to 10 kg of ${\text{H}}_{\text{2}}\text{O}$  at $\text{40°C}$  neglecting heat capacity of vessel and other losses, the equilibrium temperature will be

${\text{m}}_{l_{\text{1}}}=10\text{kg},{\text{m}}_{l_{\text{2}}}=5\text{kg}$

${\text{T}}_{l_{\text{1}}}=40°C,{\text{T}}_{l_{\text{2}}}=10°\text{C,}\text{T}=?$

We know that

${\text{Q}}_{\text{L}}={\text{Q}}_{\text{g}}$

Heat lost by hot body = heat gained by cold body

${\text{m}}_{l_{\text{1}}}{\text{s}}_{\text{w}}\left(40-\text{T}\right)={\text{m}}_{l_{\text{2}}}{\text{s}}_{\text{w}}\left(\text{T}-10\right)$

$\Rightarrow 10\times 1\left(40-\text{T}\right)=5\times 1\left(\text{T}-10\right)$

$\Rightarrow 80-2\text{T}=\text{T}-10$

$\therefore 3\text{T}=90$

$\text{T}=30°\text{C}$