A block of ice of mass 120 g at temperature $0^{\circ}\mathrm{C}$ is put in 300 gm of water at $25^{\circ}\mathrm{C}$. The xg of ice melts as the temperature of the water reaches $0^{\circ}\mathrm{C}$. The value of x is [Use: Specific heat capacity of water $=4200\,\mathrm{Jkg}^{-1}\mathrm{K}^{-1}$, Latent heat of ice $=3.5\times 10^{5}\,\mathrm{Jkg}^{-1}$]

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Heat & Thermodynamics

A block of ice of mass 120 g at temperature

0^{\circ} C

is put in 300 gm of water at

2 5^{\circ} C

. The xg of ice melts as the temperature of the water reaches

0^{\circ} C

. The value of x is
[Use: Specific heat capacity of water

= 4200 Jkg^{- 1} K^{- 1}

, Latent heat of ice

= 3.5 \times 1 0^{5} Jkg^{- 1}

]

Answer: 90

Solution

Heat released by water cooling from

2 5^{\circ} C

0^{\circ} C

Q = m c Δ T = 0.3 \times 4200 \times 25 = 31500 J

This heat melts mass

m

of ice:

m \times L = 31500

m \times 3.5 \times 1 0^{5} = 31500

m = \frac{31500}{3.5 \times 1 0 ^{5}} = 0.09 kg = 90 g

Therefore,

x = 90

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Integer answer

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