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JEE Mains · Physics · STD 11 - 10.1, thermonetry,thermal expansion and calorimetry

પાણીની વિશિષ્ટ ઉષ્મા \(=4200\, J\, kg ^{-1}\, K ^{-1}\) અને બરફની ગલનગુપ્ત ઉષ્મા \(=3.4 \times 10^{5}\, J\, kg ^{-1}\) છે \(0^{\circ} C\) તાપમાને રહેલ \(100\) ગ્રામ બરફને \(25^{\circ} C\) તાપમાને રહેલ \(200\, g\) પાણીમાં મૂકવામાં આવે છે. પાણીનું તાપમાન \(0^{\circ} C\) થાય ત્યાં સુધીમાં કેટલા ગ્રામ બરફ પીગળ્યો હશે?

A \(61.7\)
B \(63.8\)
C \(69.3\)
D \(64.6\)

Verified Solution

Answer & Solution

Correct Answer

(A) \(61.7\)

Step-by-step Solution

Detailed explanation

Here the water will provide heat for ice to melt therefore \(m_{w} s_{w} \Delta \theta=m_{i c e} L_{i c e}\) \(m _{ ice }=\frac{0.2 \times 4200 \times 25}{3.4 \times 10^{5}}\) \(=0.0617 \,kg\) \(=61.7\, gm\) Remaining ice will remain un-melted

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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