When water of mass 70g and temperature `50^(@)C ` is added to water of mass 30g, the maximum temperature of the mixutre is found to be `41^(@)C`. Find the temperature of water of mass 30g before hot water was added to it.

Explanation

Data `: m_(1) = 70g,T_(1) = 50^(@)C , m_(2) = 30g , T= 41^(@) C, T_(2) = ?`
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_(2) T_(2) = ( m_(1)+m_(2))T - m_(1)T_(1)`
`:. T_(2) = ((m_(1)+m_(2))T-m_(1)T_(1))/(m_(2))`
`= ( (70g + 30g ) 41^(@)C - 70g xx 50^(@)C)/(30g)`
`= ((100 xx 41 - 70 xx 50)/(30)) .^(@)C `
`= ((410 - 350)/( 3)) .^(@)C`
`= ( 60)/(3) .^(@)C `
`= 20^(@)C `
This is the required temperature.