200 g of water at $30°\text{C}$ is mixed with an equal amount of water at $60°\text{C}$ . Then resultant temperature of the mixture is

${\text{m}}_1=200\text{g,}{\text{t}}_1=30°\text{C,}{\text{t}}_{\text{2}}=60°\text{C,}{\text{m}}_2=200$

$\text{T}=\text{?}$

We know by principle of caloriemetry

${\text{Q}}_{\text{Lost}}={\text{Q}}_{\text{gained}}$

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

$\Rightarrow \text{200×1×}\left(60-T\right)=200\times 1\times \left(T-30\right)$

$\Rightarrow 60-\text{T}=\text{T}-30$

$\therefore 2\text{T}=90$

$\text{T}=45°\text{C}$