A liquid at $$30$$ºC is poured very slowly into a Calorimeter that is at temperature of $$110$$ºC. The boiling temperature of the liquid is $$80$$ºC. It is found that the first $$5gm$$ of the liquid completely evaporates. After pouring another $$80gm$$ of the liquid the equilibrium temperature is found to be $$50$$ºC. The ratio of the Latent heat of the liquid to its specific heat will be $$___$$ºC.[ Neglect the heat exchange with surrounding and assume calorimeter is open]

Question

Infinity Learn · Accepted Answer

Step $$1$$:Heat lost by calorimeter  $$=mcalScal$$×$$110-80Heat$$ gained by $$liquid=5$$×$$Sliq.80-30+5Lf$$     $$mCalSCal30=Sliq250+5Lf___1Step$$ $$2$$: Heat lost by calorimeter $$=$$  mcalScal×$$80-50Heat$$ gained by liq $$=$$    $$80$$×Sliq×$$50-30$$      mCalSCal×$$30=80$$ $$Sliq20___21600Sliq$$−$$250Sliq=5Lf$$⇒$$1350Sliq=5Lf$$⇒$$LfSliq=13505=270$$

Calorimetry

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Calorimetry

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Step 1:Heat lost by calorimeter =mcalScal×110-80Heat gained by liquid=5×Sliq.80-30+5Lf mCalSCal30=Sliq250+5Lf___1Step 2: Heat lost by calorimeter = mcalScal×80-50Heat gained by liq = 80×Sliq×50-30 mCalSCal×30=80 Sliq20___21600Sliq−250Sliq=5Lf⇒1350Sliq=5Lf⇒LfSliq=13505=270

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