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

Step-by-step solution

Given: ∆ Q = 184 × 10 3 J , m = 0 . 600 kg at - 12 ° C , S = 2222 . 3 J kg – 1 ° C – 1 and L = 336 × 10 3 J kg - 1 Heat required to convert ice from – 12 ° C to ice at 0 ° C will be, Q 1 = 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 , Q 2 = 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