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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
 
begin mathsize 14px style fraction numerator d Q over denominator d t space end fraction space equals space left parenthesis space m space cross times space C subscript p space plus space h space right parenthesis cross times fraction numerator d T over denominator d t end fraction end style  .......................................(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
 
begin mathsize 14px style fraction numerator d T over denominator d t end fraction space equals space k space left parenthesis space T space minus space T subscript o subscript space end subscript right parenthesis end style
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
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Asked by swayamagarwal2114 | 25 Jul, 2022, 02:34: PM
ANSWERED BY EXPERT
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