The time taken for a calorimeter containing 75g of water at `62^(@)C` to cool to `58^(@)C` is 9 minutes. When the calorimeter contains 105g of water, it takes 12 minutes to cool from `62^(@)C " to " 58^(@)C`. The water equivalent of the calorimeter is

Options

A. 10g
B. 15 g
C. 20g
D. 30g

Answer

B

Explanation

Heat lost `= (w + m) C_(v) dT`
where, w = water equivalent of calorimeter,
m = mass of water,
`C_(v)`= specific heat of water and dT = temperature difference `=T_(2) - T_(1)`. Time taken to cool down is proportional to amount of heat lost.
Case I 9 minutes
`prop (w + 75) C_(v) xx (62^(@) - 58^(@))`...(i)
Case II 12 minutes
`prop (w + 105) C_(v) xx (62^(@) - 58^(@))`...(ii)
Dividing Eq. (i) by (ii), we get
`(9)/(12) = ((75 + w) C_(v) xx 4^(@))/((1.05 + w) C_(v) xx 4^(@))`
`3 xx (105 + w) = 4(75 + w)`
or `4w -3w = 315g - 300g`
w=15g