The specific heat of water =4200 J kg -1 K -1 and the latent heat of ice =3.4 × 105 J kg -1. 100 grams of ice at 0° C is placed in 200 g of water at 25° C. The amount of ice that will melt as the temperature of water reaches 0° C is close to (in grams) :

Question

The specific heat of water =4200 J kg -1 K -1 and the latent heat of ice =3.4 × 105 J kg -1. 100 grams of ice at 0° C is placed in 200 g of water at 25° C. The amount of ice that will melt as the temperature of water reaches 0° C is close to (in grams) : (A) 61.7 (B) 63.8 (C) 69.3 (D) 64.6

Tardigrade Admin · Accepted Answer

Here the water will provide heat for ice to melt
therefore

m_{w} s_{w} Δ θ = m_{i ce} L_{i ce}

m_{i ce} = \frac{0.2 \times 4200 \times 25}{3.4 \times 1 0 ^{5}}

= 0.0617 k g

= 61.7 g m

Remaining ice will remain un-melted
so correct answer is 1

Q. The specific heat of water =4200Jkg−1K−1 and the latent heat of ice =3.4×105Jkg−1. 100 grams of ice at 0∘C is placed in 200g of water at 25∘C. The amount of ice that will melt as the temperature of water reaches 0∘C is close to (in grams) :

61.7

63.8

69.3

64.6

Here the water will provide heat for ice to melt therefore mw​sw​Δθ=mice​Lice​ mice​=3.4×1050.2×4200×25​ =0.0617kg =61.7gm Remaining ice will remain un-melted so correct answer is 1

Here the water will provide heat for ice to melt
therefore
$m_{w} s_{w} Δ θ = m_{i ce} L_{i ce}$
$m_{i ce} = \frac{0.2 \times 4200 \times 25}{3.4 \times 1 0 ^{5}}$
$= 0.0617 k g$
$= 61.7 g m$
Remaining ice will remain un-melted
so correct answer is 1