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प्रश्न
A mass of 50 g of a certain metal at 150° C is immersed in 100 g of water at 11° C. The final temperature is 20° C. Calculate the specific heat capacity of the metal. Assume that the specific heat capacity of water is 4.2 J g-1 K-1.
उत्तर
Heat liberated by metal = m × s × Δ t
= 50 × s × (150 – 20)
Heat absorbed by water = mw × sw × Δt
= 100 × 4.2 × (20 – 11)
Heat energy lost = heat energy gained
⇒ 50 × s × (150 – 20) = 100 × 4.2 × (20 – 11)
⇒ 50 × s × 130 = 100 × 4.2 × 9
⇒ s = `(100 × 4.2 × 9)/(50 xx 130)`
s = `37.8/65`
S = 0.58 J g-1 K-1.
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