The specific heat of water $=4200 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$ and the latent heat of ice $=3.4 \times 10^{5} \mathrm{~J} \mathrm{~kg}^{-1}$. 100 grams of ice at $0^{\circ} \mathrm{C}$ is placed in 200 g of water at $25^{\circ} \mathrm{C}$. The amount of ice that will melt as the temperature of water reaches $0^{\circ} \mathrm{C}$ is close to (in grams):

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Heat & Thermodynamics

The specific heat of water

= 4200 J kg^{- 1} K^{- 1}

and the latent heat of ice

= 3.4 \times 1 0^{5} J kg^{- 1}

. 100 grams of ice at

0^{\circ} C

is placed in 200 g of water at

2 5^{\circ} C

. The amount of ice that will melt as the temperature of water reaches

0^{\circ} C

is close to (in grams):

(A)64.6
(B)61.7✓ Correct
(C)69.3
(D)63.8

Answer: (B)

Solution

The warm water cools to

0^{\circ}

C and the heat it gives up melts some of the ice. Heat released by the water is

Q = m_{w} c_{w} Δ T = 0.2 \times 4200 \times 25 = 21000

J. Each kilogram of ice needs

L = 3.4 \times 1 0^{5}

J to melt, so the mass melted is

m = \frac{21000}{3.4 \times 1 0 ^{5}} = 0.0617

kg, about

61.7

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Single correct option

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