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# JEE Class main Answered

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Asked by swayamagarwal2114 | 25 Jul, 2022, 02:34: PM Expert Answer
Mass of liquid in calorimeter = volume × density = 100 × 0.88 = 88 gram

electrical power of coil = voltage × current = 6.3 × 2 = 12.6 W = 12.6 J/s

while heating, at steady state,  we have .......................................(1)

where dQ/dt is heat absorbed by liquid and calorimeter that is same as electrical power of coil.

m is mass of liquid, Cp is specific heat,  h is water equivalent of calorimeter and
dT/dt is rate of change of temeperature

( 88 × 10-3 × Cp + 13 × 10-3 ) × (3.6/60) = 12.6

From above expression, we get Cp = 2386 J/kg

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By Newtons law of cooling , we have where k is constant and To is room temperature

From above expression we get , k = (3.6/60) / ( 55 - 10) = 1.333 × 10-3 J s-1 K-1

if room temeperature is 26o C , then rate of loss of heat by cooling is calcultaed as

(dT/dt ) = 1.333 × 10-3 × ( 55 - 26 ) = 3.867 × 10-2 J/s

Hence electrical power required for heating coil to get steady state is determined from eqn.(1)
= ( 88 × 10-3 × 2386 + 13 × 10-3 ) × 3.867 × 10-2 = 8.12 W

Answered by Thiyagarajan K | 25 Jul, 2022, 09:14: PM
JEE main - Physics
. Asked by swayamagarwal2114 | 25 Jul, 2022, 02:34: PM ANSWERED BY EXPERT
JEE main - Physics
Asked by familynhmpns | 04 Jul, 2022, 06:34: PM ANSWERED BY EXPERT