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) = 70` g , `T_(1) = 50^(@)C , m_(2) = 30 g , T = 41^(@)C , T_(2)` = ?
According to the principle of heat exchange ,heat lost by hot water = heat gained by cold water
` therefore m_(1)c (T_(1) - T) = m_(2) c (T - T_(2))`
`therefore m_1T_(1) - m_1 T = m_(2) T - m_(2) T_(2)`
`therefore m_(2) T_(2) = (m_(1) + m_(2)) T - m_(1) T_(1)`
`therefore T_(2) = ((m_(1) + m_2) T - m_(1) T_1)/(m_(2))`
`= ((70 g + 30 g ) 41^(@) C - 70 g xx 50^(@)C)/(30 g )`
= `((100 xx 41 - 70 xx 50)/(30)) ""^(@)C`
`= ((410 - 350)/(3)) ""^(@) C = (60)/(3) ""^(@)C = 20 ""^(@)C`
This is the required temperature .