JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Heat & Thermodynamics

Heat energy of 184 kJ is given to ice of mass 600 g at

- 1 2^{\circ} C

. Specific heat of ice is

2222.3 J kg^{- 1 \circ} C^{- 1}

and latent heat of ice in

336 kJ/kg^{- 1}

(A) Final temperature of system will be

0^{\circ} C

.
(B) Final temperature of the system will be greater than

0^{\circ} C

.
(C) The final system will have a mixture of ice and water in the ratio of

5 : 1

.
(D) The final system will have a mixture of ice and water in the ratio of

1 : 5

.
(E) The final system will have water only.
Choose the correct answer from the options given below:

Answer: (A)

Solution

Bringing the ice from

- 1 2^{\circ}

C to the melting point needs

Q_{1} = m c Δ T = 0.6 \times 2222.3 \times 12 \approx 1.6 \times 1 0^{4}

J, leaving

1.84 \times 1 0^{5} - 1.6 \times 1 0^{4} = 1.68 \times 1 0^{5}

J of the supplied heat. Melting all

600

g would require

m L = 0.6 \times 336000 = 2.016 \times 1 0^{5}

J, which is more than remains, so only part of the ice melts and the mixture rests at

0^{\circ}

C. The mass melted is

\frac{1.68 \times 1 0 ^{5}}{3.36 \times 1 0 ^{5}} = 0.5

kg, leaving

0.1

kg of ice, so the final state is ice and water at

0^{\circ}

C in the ratio

0.1 : 0.5 = 1 : 5

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Single correct option