Heat energy of 184   kJ is given to ice of mass 600   g at - 12   ° C , Specific heat of…

Heat energy of 184 kJ is given to ice of mass 600 g at -12 °C, Specific heat of ice is 2222.3 J kg–1 °C–1 and latent heat of ice is 336 kJ kg–1.

(A) Final temperature of system will be 0 °C

(B) Final temperature of the system will be greater than 0 °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 :

  1. A and D only
  2. B and D only
  3. A and E only
  4. A and C only

Solution

Given:

∆Q=184×103 J,

m=0.600 kg at -12 °C,

S=2222.3 J kg–1 °C–1 and

L=336×103J kg-1

Heat required to convert ice from –12 °C to ice at 0 °C will be, 

Q1=0.600×2222.3×12=16000.56 J

Remaining heat Q=184000-16000.56

= 167999.44 J

Heat required to melt ice completely at 0 °C,

Q2 = 0.600 × 336000 = 201600 J needed which is more than Q, so 100% ice is not melted.

For amount of melted ice, we can write

167999.44=m×336000⇒m= 0.4999 kg

∴ mass of water =0.4999 kg

Mass of ice remaining = 0.1001

∴ Ratio =0.10010.4999≈1:5

Asked in: JEE Main 2023 (29 Jan Shift 2)

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