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Question

A piece of brass of mass 200g at 100°C is placed in 400g of turpentine oil, contained in a copper calorimeter of mass 50g at15°C. The final temperature of the mixture is23°C. Find the specific heat capacity of turpentine oil.


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Solution

Step 1: Organised table of the given data

SubstanceMassS.H.CInitial TemperatureFinal Temperature=23°C
Brass200g=2001000kg370JKg-1K-1100°CθF=100-23=77°C
Oil400g=4001000kg?15°CθR=23-15=8°C
Calorimeter50g=501000kg390JKg-1K-115°C

Step 2: Finding the specific heat capacity of turpentine oil

Let the specific heat capacity of the oil is Ct.

Heat gained by turpentine + calorimeter =4001000×Ct+501000×C(23-15)

=25Ct+120×390×8=25Ct+3928=165Ct+156J

Now, heat loss of brass=mc(dT)

So, the heat lost is

0.2×370×77°=5698J

Since, Heat gained=Heat lost

165Ct+156=5698165Ct=5698156165Ct=5542Ct=5542×516Ct=1731.8JKg1K1

Hence, the sp heat capacity of turpentine oil is 1731.8JKg1K1.


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