हिंदी

A solid of mass 50 g at 150°C is placed in 100 g of water at 11°C when the final temperature recorded is 20°C. Find the specific heat capacity of the solid.

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प्रश्न

A solid of mass 50 g at 150°C is placed in 100 g of water at 11°C when the final temperature recorded is 20°C. Find the specific heat capacity of the solid. (specific heat capacity of water = 4.2 J/g °C)

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उत्तर

Mass of the solid, ms = 50 g

Initial temperature of the solid, ts = 150°C

Mass of water, mw = 100 g

Temperature of water, tw = 11°C

Final temperature of the mixture, t = 20°C

According to the principle of calorimetry

Heat gained by water = Heat lost by the solid

`:.m_wc_w(t - t_w) = m_sc_s(t_s - t)`

∴ 100 × 4.2 × (20 - 11) = 50 × cs × (150 - 20)

3780 = 6500 cs

`c_s = 3780/6500`

`∴ c_s = 0.58` J/g °C

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अध्याय 10: Calorimetry - NUMERICAL PROBLEMS ON SPECIFIC HEAT CAPACITY [पृष्ठ २३९]

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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 3. | पृष्ठ २३९
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