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When 300 J of heat is added to 25 gm of sample of a material its temperature rises from 25°C to 45°C. The thermal capacity of the sample and specific heat of the material are respectively given by:

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$600 J /^{0} C, 15 J /^{0} C - kg$

none of these

$15 J / ° C, 600 J / Kg - ° C$

$150 J /^{0} C, 60 J / Kg -^{0} C$

answer is A.

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Detailed Solution

(i) Thermal energy

$TC = mc = \frac{Q}{∆ T} = \frac{300}{45 - 25} = 15 J /^{0} C$

(ii) Specific heat is nothing but thermal capacity per unit mass

$C = \frac{mc}{m} = \frac{15}{25 \times 10^{- 3}} = 600 J / kg -^{0} C$

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