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100 ml of water at 20^(@)C and 100 ml of...

`100 ml` of water at `20^(@)C` and `100 ml` of water at `40^(@)C` are mixed in calorimeter until constant temperature reached. Now temperature of the mixture is `28^(@)C`. Water equivalent of calorimeter is

A

50J

B

104.5J

C

`-24.2J`

D

209J

Text Solution

Verified by Experts

The correct Answer is:
D
`W=[(m_(2)(t_(2)-t_(3)))/(t_(3)-t_(1))-m_(1)]`
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