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CBSE Class 12-science Answered

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Asked by jain.pradeep | 29 Apr, 2019, 09:10: PM
answered-by-expert Expert Answer
Qn. 31:- the curve shows it is almost reaching the surrounding temperature 20 °C and change in temperature
                 keep on decreasing and it is very less at   4  hour.  Hence after 8 hours of cooling, temperature will be very near to 20 °C
 
Qn. 32:-
Energy loss  ΔE for change in temperature ΔT is given by  ΔE = m×Cp×ΔT ,  
where m is mass of water and Cp is specific heat of water

if  ΔT = 1°C   ,  then Energy loss per °C,    m×Cp = 8×105 J /°C .................(1)
 
Rate of change of energy loss with repect to time,  (ΔE/Δt) = m×Cp×( ΔT/Δt) = 16×106 J/hour  ................(2)
 
from eqn.(1) and (2),  Rate of temperature falling,  ( ΔT/Δt) = 16×106 / ( 8×105)  = 20 °C/hour
 
Answered by Thiyagarajan K | 30 Apr, 2019, 10:45: AM
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