If water of mass 80g and temperature `45^(@)C ` is mixed with water of mass 20g and temperature `30^(@)C ` , what will be maximum temperature of the mixture?

Explanation

Data `: m _(1) = 80 g, T_(1) = 45^(@)C, m_(2) = 20 g ,T_(2) = 30^(@) C , T=?`
According to the principle of heat exchange, heat lost by hot water = heat gained by cold water
`:. m _(1)c( T_(1)-T) = m_(2)c(T-T_(2))`
`:. m_(1)T_(1) - m_(1)T= m_(2)T - m_(2)T_(2)`
`:. m_(1)T_(1)+m_(2)T_(2) = ( m_(1)+m_(2))T`
`:.` Maximum temperature of the mixture.
`T= ( m_(1)T_(1)+m_(2)T_(2))/(m_(1)+m_(2))`
`= ( 80gxx 45^(@)C + 20 g xx30^(@)C)/(80g + 20 g)`
`= ((80 xx 45)/(100) + ( 20 xx 30)/( 100)).^(@)C `
` = ( 36+ 6).^(@)C`
`= 42^(@)C`