A block of ice of mass 120 g at temperature 0° C is put in 300 gm of water at 25° C. The xg of ice melts as the temperature of the water reaches 0° C. The value of x is [Use: Specific heat capacity of water =4200 Jkg -1 K -1, Latent heat of ice .=3.5 × 105 Jkg -1]

Question

A block of ice of mass 120 g at temperature 0° C is put in 300 gm of water at 25° C. The xg of ice melts as the temperature of the water reaches 0° C. The value of x is [Use: Specific heat capacity of water =4200 Jkg -1 K -1, Latent heat of ice .=3.5 × 105 Jkg -1] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Energy released by water =0.3 × 25 × 4200=31500 J let m kg ice melts m × 3.5 × 105=31500 m =(31500 × 10-5/3.5)=9000 × 10-5 m =0.09 kg =90 gm x =90

Q. A block of ice of mass 120g at temperature 0∘C is put in 300gm of water at 25∘C. The xg of ice melts as the temperature of the water reaches 0∘C. The value of x is [Use: Specific heat capacity of water =4200Jkg−1K−1, Latent heat of ice =3.5×105Jkg−1]

Energy released by water =0.3×25×4200=31500J let mkg ice melts m×3.5×105=31500 m=3.531500×10−5​=9000×10−5 m=0.09kg=90gm x=90

Energy released by water $= 0.3 \times 25 \times 4200 = 31500 J$
let $mk g$ ice melts
$m \times 3.5 \times 1 0^{5} = 31500$
$m = \frac{31500 \times 1 0 ^{- 5}}{3.5} = 9000 \times 1 0^{- 5}$
$m = 0.09 k g = 90 g m$
$x = 90$