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Standard VIII

Physics

Expansion of Liquids

An immersion ...

Question

An immersion heater, in an insulated vessel of negligible heat capacity brings $10 g$ of water to the boiling point from $16^{\circ} C$ in $7 m i n$ . ThenThe water is replaced by $20 g$ of alcohol, which is heated from $16^{\circ} C$ to the boiling point of $78^{\circ} C$ in $6 m i n 12 s$ whereas $30 g$ are vapourized in $5 m i n 6 s$ . The specific heat of alcohol is

0.6 J / k g^{\circ} C

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0.6 c a l / g^{\circ} C

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0.6 c a l / k g^{\circ} C

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6 J / k g^{\circ} C

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Solution

The correct option is A $0.6 J / k g^{\circ} C$
Time= t= $7 m i n = 7 \times 60 s = 420 s$
Hence, power of heater=P=heat supplied/time $= m C Δ T / t = 10 \times 1 \times 84 / 420 = 2 c a l / s$
Hence heat supplied to 20g alcohol= $P \times t i m e = 2 \times 372 c a l = 744 c a l = m C Δ T = 20 \times 62 \times C = 1240 C$
$\Rightarrow C = \frac{744}{1240} = 0.6 c a l / g K$

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The correct option is A 0.6J/kg∘CTime= t= 7min=7×60s=420sHence, power of heater=P=heat supplied/time=mCΔT/t=10×1×84/420=2cal/sHence heat supplied to 20g alcohol=P×time=2×372 cal=744 cal=mCΔT=20×62×C=1240C⇒C=7441240=0.6 cal/gK