Verified Solution • 2025–26 Session moderate

QJEEM23Ph835b: JEE Main 2023 Physics - Class 11 JEEM23Ph835b Solution: Heat energy of 184 kJ is given to ice of mass 600 g at .…

Heat energy of 184 kJ is given to ice of mass 600 g at −12∘C-12^\circ \mathrm{C} . Specific heat of ice is 2222.3 J kg−1∘ C−1\mathrm{2222.3~J~kg^{-1^\circ}~C^{-1}} and latent heat of ice in 336 kJ/kg−1\mathrm{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 :

Multiple Choice Options

  • A A and E only
  • B B and D only
  • C A and C only
  • D A and D only

Step-by-Step Verified Solution

Heat required to raise the temperature of ice to 0 ∘^\circ C is

=601000(2222.3)(12)= {{60} \over {1000}}(2222.3)(12)

=16000.5= 16000.5 J

≈16\approx 16 kJ

Heat required to melt ice completely

=(6001000)(336)= \left( {{{600} \over {1000}}} \right)(336) kJ

=201.6= 201.6 kJ

Energy left =(184−16)=168= (184 - 16) = 168 kJ

∴\therefore Partial ice will melt

∴\therefore 168=(mice melted)336168 = ({m_{ice\,melted}})336

0.50.5 kg =(mice melted)= ({m_{ice\,melted}})

∴\therefore mice:mwater=1:5{m_{ice}}:{m_{water}} = 1:5

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