JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Heat & Thermodynamics

Ice at

- 2 0^{\circ} C

is added to 50 g of water at

4 0^{\circ} C

. When the temperature of the mixture reaches

0^{\circ} C

, it is found that 20 g of ice is still unmelted. The amount of ice added to the water was close to (Specific heat of water

= 4.2 J / g /^{\circ} C

)
Heat of fusion of water at

0^{\circ} C = 334 J / g

)

Answer: (D)

Solution

Let

m

grams of ice at

- 2 0^{\circ}

C be added. The warm water cools from

4 0^{\circ}

C to

0^{\circ}

C, releasing

50 \times 4.2 \times 40 = 8400 J

. The ice first warms to

0^{\circ}

C, taking

m \times 0.5 \times 20 = 10 m J

, and then part of it melts; since

20

g remains solid, only

(m - 20)

g melts, needing

334 (m - 20) J

. Equating heat gained to heat lost,

10 m + 334 (m - 20) = 8400

, so

344 m = 15080

and

m \approx 43.8 g

, closest to

40 g

JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Single correct option