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

એક પ્રયોગમાં \(0.20\, kg\)ના એલ્યુમિનિયમના સળિયાને \(150\,^oC\) સુધી ગરમ કરેલ છે. તેને \(0.025\, kg\). કેલોરીમીટરના પાણી સમતુલ્ય માં \(27\,^oC\) તાપમાને રહેલ \(150\, cc\) કદના પાણીમાં મૂકવામાં આવે છે.તંત્રનું અંતિમ તાપમાન \(40\,^oC\) છે. એલ્યુમિનિયમની વિશિષ્ટ ઉષ્મા \(J/kg\,-\,^oC\) માં કેટલી હશે?( \(4.2\, Joule= 1\, calorie\))

A \(378\)
B \(315\)
C \(476\)
D \(434\)

Verified Solution

Answer & Solution

Correct Answer

(D) \(434\)

Step-by-step Solution

Detailed explanation

According to principle of calorimetry, \({Q_{given}} = {Q_{used}}\) \(0.2 \times S \times \left( {150 - 40} \right) = 150 \times 1 \times \left( {40 - 27} \right) + 25\) \( \times \left( {40 - 27} \right)\) \(0.2 \times S \times 110 = 150 \times 13 + 25 \times 13\) Specificheat…

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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