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 = 4200 J/kg^oC , Specific latent heat of fusion of ice = 336 x 10³ J/kg.

Q. X g of ice at

0^{o}

C is added to

340

g of water at

20^{o}

C. The final temperature of the resultant mixture is

5^{o}

C. The value of X (in g) is closest to. [Heat of fusion of ice

= 333 J / g

; Specific heat of water

= 4.184 J / (g . K)

Join BYJU'S Learning Program

Ice at −20oC is dropped into 0.25kg of the water at 20oC. The final temperature of the system is 0oC, when all the ice is melted. Find approximate mass of ice in the water (Specific heat of ice =2000Jkg−1K−1,L=33.4×104Jkg−1).

Let the required amount of ice be mKgHeat gain = Heat lossm×2000×[0−(−20)]+m×33.4×104=0.25×4200×20⇒m=0.25×4200×20(2000×20)+33.4×104=0.056Kg

Ice at $- 20^{o}$ C is dropped into $0.25$ kg of the water at $20^{o}$ C. The final temperature of the system is $0^{o}$ C, when all the ice is melted. Find approximate mass of ice in the water (Specific heat of ice $= 2000 J k g^{- 1} K^{- 1}, L = 33.4 \times 10^{4} J k g^{- 1}$ ).

Let the required amount of ice be $m K g$
Heat gain = Heat loss
$m \times 2000 \times [0 - (- 20)] + m \times 33.4 \times 10^{4} = 0.25 \times 4200 \times 20 \Rightarrow m = \frac{0.25 \times 4200 \times 20}{(2000 \times 20) + 33.4 \times 10^{4}} = 0.056 K g$