Question

plz ans my question fast I have exams​

Answer 1

$\huge\underline{\underline{\bf \green{Question-}}}$

A piece of ice of mass 40 g is dropped into 200 g of water at 50°C. Calculate the final temperature of water after all the ice has melted. Specific heat capacity of water ${\sf 4200\:Jkg^{-1}K^{-1}}$ , Specific latent heat of fusion of ice = 336 x 10³${\sf Jkg^{-1}}$.

$\huge\underline{\underline{\bf \green{Solution-}}}$

$\large\underline{\underline{\sf Given:}}$

• Mass of ice = 40 g
• Mass of water = 200 g
• Temperature = 50°C

$\large\underline{\underline{\sf To\:Find:}}$

• Final temperature of water after all the ice has melted

We know that ,

✪$\large{\boxed{\bf \blue{Q=mc\triangle T}}}$

Q = heat

m = mass

c = Specific heat

Let final temperature be = x

✪ Heat absorbed by ice = Heat diffused ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀by water

Heat absorbed by ice at latent heat ⎯

$\implies{\sf \dfrac{40×336×1000}{1000}}$

$\implies{\sf \orange{13440J} }$

Heat absorbed by ice when heating to x ⎯

$\implies{\sf mc\triangle T}$

$\implies{\sf \dfrac{40×4200×(x-0)}{1000}}$

$\implies{\sf \orange{ 168x}}$

Heat diffused by water ⎯

$\implies{\sf \dfrac{200}{1000}×4200(50-x)}$

$\implies{\sf \orange{ 840(50-x)}}$

$\implies{\sf 168x+13440=840(50-x)}$

$\implies{\sf 168x+13440=42000-840x }$

$\implies{\sf 168x+840x=42000-13440}$

$\implies{\sf 1008x=28560}$

$\implies{\bf \red{Final\: Temperature=28.3°C} }$

$\huge\underline{\underline{\bf \green{Answer-}}}$

Final temperature ${\bf \red{28.3°C}}$