10g of ice at$0°\text{C}$ is mixed with 100g of water at$50°\text{C}$ what is resultant temperature

given mass of ice =${\text{m}}_1=10\text{g}$  

Initial temperature $=\text{0°C}$

Mass of water ${\text{m}}_{\text{2}}=\text{100g}$

Initial temperature$=\text{50°C}$

${\text{m}}_{\text{1}}\text{L}+{\text{m}}_{\text{1}}\text{s}\Delta {\text{t}}_{\text{1}}={\text{m}}_{\text{2}}\text{s}\Delta {\text{t}}_{\text{2}}$

$\Rightarrow 10\times 80+10\times 1\times \left(\text{T}-0\right)=100\times 1\times \left(50-T\right)$

$\Rightarrow 110\text{T}=4200\Rightarrow \text{T}=38.2°\text{C}$