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A piece of brass of mass 200 g and at 100°C, is placed in 400 g of turpentine oil, contained in a copper calorimeter of mass 50 g at 15°C. The final temperature recorded is 23°C.

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Question

A piece of brass of mass 200 g and at 100°C, is placed in 400 g of turpentine oil, contained in a copper calorimeter of mass 50 g at 15°C. The final temperature recorded is 23°C. Find the sp. heat capacity of turpentine oil.

[S.H.C. for brass = 370 Jkg−1K−1; S.H.C. of copper = 390 Jkg−1K−1]

Numerical
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Solution

Using SI units:

brass mass mb = 0.2 kg

turpentine mass mt = 0.4 kg

calorimeter mass mc = 0.05 kg

Initial temperatures:

brass 100°C

turpentine and calorimeter = 15°C

final temperature θ = 23°C

Fall in temperature of brass:

θF = 100 − 23 = 77°C

Rise in temperature of turpentine and calorimeter:

θR = 23 − 15 = 8°C

Heat lost by brass:

Qb = mbcbθF

= 0.2 × 370 × 77

= 5698 J

Heat gained by calorimeter:

Qc = mcccθR

= 0.05 × 390 × 8

= 156 J

Heat gained by turpentine:

Qt = mtctθR

= 0.4 × ct × 8

= 3.2ct J

By the principle of calorimetry,

Qb = Qt + Qc

5698 = 3.2ct + 156

3.2ct = 5542

$$c_t=\frac{5542}{3.2}$$

= 1731.875

≈ 1731.9 J kg−1 K−1

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Chapter 10: Calorimetry - NUMERICAL PROBLEMS ON SPECIFIC HEAT CAPACITY [Page 240]

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Goyal Brothers Prakashan A New Approach to ICSE Physics [English] Class 10
Chapter 10 Calorimetry
NUMERICAL PROBLEMS ON SPECIFIC HEAT CAPACITY | Q 2. | Page 240
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