An amount of water of mass 20 g at `0^(@)C` is mixed with 40 g of water at `10^(@)C` . Final temperature of mixture is

Options

A. `20^(@)C`
B. `6.67^(@)C`
C. `5^(@)C`
D. `0^(@)C`

Answer

B

Explanation

Resultant temperature , `t_(R) = (m_(1) s.t_(1) + m_(2) st_(2))/(m_(1)s +m_(2)s)`
`implies t_(R) = ((20)(0)+(40)(10))/((20)+(40))`
`therefore t_(R) =(400)/(60)`
=`6.67^(@)C`