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

Options

A. 5 g
B. 10 g
C. 25 g
D. 50 g

Answer

B

Explanation

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`