Q.

A liquid at  30°C  is poured very slowly into a calorimeter that is at temperature of 110°C  . The boiling temperature of the liquid is  80°C  . It is found that the first 5 g of the liquid completely evaporates. After pouring another 80 g of the liquid the equilibrium temperature is found to be  50°C . The ratio of the latent heat of the liquid to its specific heat will be  x×90°C ,  then   x=   ________ [Neglect the heat exchange with surrounding]

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answer is 3.

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Detailed Solution

Let  mcal=   mass of calorimeter
 Scal= specific heat of calorimeter
Sliq=  specific heat of liquid
L=latent heat of liquid
Step-1:
Heat lost by calorimeter  =mcalscal×30
Heat gained by liquid  =5×sliq.50+mL
 mliqscal30=sliq50+5Lf
Step-2:
 Heat lost by calorimeter  mcalscal×30
Heat lost by liquid  =80×sliq×20
 mcalscal×30=80sliq20  

   1600sliq250sliq=5Lf 

  1350sliq=5Lf 

 Lfsliq=13505=270
 
 
 
 

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