हिंदी

Heat energy is given to 150 g of water, such that its temperature rises by 20 K. When the same amount of heat energy is given to a liquid X of mass 100 g its temperature rises by 80 K.

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प्रश्न

Heat energy is given to 150 g of water, such that its temperature rises by 20 K. When the same amount of heat energy is given to a liquid X of mass 100 g its temperature rises by 80 K. Calculate (a) heat energy given to water (b) the specific heat capacity of liquid X.

[Take sp. heat capacity of water = 4.2 J g−1 K−1]

संख्यात्मक
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उत्तर

(a) Heat energy given to water = mcθR

 = 150 g × 4.2 J g−1 K−1 × 20 K

= 12600 J

(b) Heat energy given to liquid X = 12600 J

Now, Heat energy given to liquid X = mcθR

12600 J = 100 g × с × 80 K

\[{c} =\frac{12600 \mathrm{~J}}{100 \mathrm{~g} \times 80 \mathrm{~K}}\]

∴ c = 1.575 J g−1 K−1

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अध्याय 10: Calorimetry - NUMERICAL PROBLEMS ON SPECIFIC HEAT CAPACITY [पृष्ठ २४१]

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गोयल ब्रदर्स प्रकाशन A New Approach to ICSE Physics [English] Class 10
अध्याय 10 Calorimetry
NUMERICAL PROBLEMS ON SPECIFIC HEAT CAPACITY | Q 11. | पृष्ठ २४१
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