# A calorimeter contains 10 g of water at 20^(@)C. The temperature falls to 15^(@)C in 10 min. When calorimeter contains 20 g of water at 20^(@)C,

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A calorimeter contains 10 g of water at 20^(@)C. The temperature falls to 15^(@)C in 10 min. When calorimeter contains 20 g of water at 20^(@)C, it takes 15 min for the temperature to becomes 15^(@)C. The water equivalent of the calorimeter is
A. 5 g
B. 10 g
C. 25 g
D. 50 g

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Let water equivalent be x
therefore (10+x)c((DeltaT)/(DeltaT))= -k(T_(1)+T_(2))/2 - T_(0))
Where c is the specific heat of water
DeltaT = Change in temperature
Deltat = time at which temperature falls
therefore (10+x)c(20-15)/(15) = -k(20+15)/2 -T_(0) ................(i)
(20+x)c(20-15)/15 = -k((20+15)/2)= -k((20+15)/2)-T_(0)).....................(ii)
From eqs. (i) and (ii), we get
(10+x)/10 = (20+x)/15
150+15x = 200 + 10x
5x =50 rArr x=10g