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Question

500g of molten lead at its melting point is just kept in a molten state by a heater of 25W. If the heater is switched off, the temperature of lead starts falling after 7min. Calculate the specific latent heat of fusion of lead.


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Solution

Step 1: Given data

  1. The quantity of molten lead is 500g
  2. The power of the heater is 25W

Step 2: Calculating the heat given out by lead

Let the specific latent heat of lead is LJg-1.

Heat given out by lead at its melting point

=amountoflead×Specificlatentheatoflead=500×L

The heat is given by the lead

=(Powerofheater×time)=(25×7×60)J=10500J

Step 3: Calculating the specific latent heat of fusion of lead

By the principle of calorimetry

Heat given out by lead at itsmeltingpoint=Heat given out by lead500×L=10500L=10500500L=21Jg-1

Hence, the latent heat of lead is 21Jg-1.


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