# 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, C

_{p}is specific heat, h is water equivalent of calorimeter anddT/dt is rate of change of temeperature

( 88 × 10

^{-3}× C_{p}+ 13 × 10^{-3}) × (3.6/60) = 12.6From above expression, we get C

_{p}= 2386 J/kg------------------------------------------

By Newtons law of cooling , we have

where k is constant and T

_{o}is room temperatureFrom above expression we get , k = (3.6/60) / ( 55 - 10) = 1.333 × 10

^{-3}J s^{-1}K^{-1}if room temeperature is 26

^{o}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/sHence electrical power required for heating coil to get steady state is determined from eqn.(1)

= ( 88 × 10

^{-3}× 2386 + 13 × 10^{-3}) × 3.867