Verified Solution • 2025–26 Session moderate

QJEEM19Phb390: JEE Main 2019 Physics - Class 11 JEEM19Phb390 Solution: When M 1 gram of ice at –10 o C (specific heat = 0.5 cal g –1 o …

When M1 gram of ice at –10oC (specific heat = 0.5 cal g–1 oC–1 ) is added to M2 gram of water at 50C, finally no ice is left and the water is at 0°C. The value of latent heat of ice, in cal g–1 is :

Multiple Choice Options

  • A 50M2M1−5{{50{M_2}} \over {{M_1}}} - 5
  • B 50M2M1{{50{M_2}} \over {{M_1}}}
  • C 5M2M1−5{{5{M_2}} \over {{M_1}}} - 5
  • D 5M1M2−50{{5{M_1}} \over {{M_2}}} - 50

Step-by-Step Verified Solution

Heat lost = Heat gain

⇒M2×1×50=M1×0.5×10+M1.Lf\Rightarrow {M_2} \times 1 \times 50 = {M_1} \times 0.5 \times 10 + {M_1}.{L_f}

⇒Lf=50M2−5M1M1\Rightarrow {L_f} = {{50{M_2} - 5{M_1}} \over {{M_1}}}

=50M2M1−5= {{50{M_2}} \over {{M_1}}} - 5

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