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

Explanation

Law of conservation of energy : Energy can neither be created nor destroyed. It can be converted from one form into another . Thus, the total amount of energy in the universe always remains constant.
Data : `m_(1) = 80 g, T_(1) = 45^(@) C, m_(2) = 20 g, T_(2)= 30^(@)C, T = ?`
Accordin 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))`
` ( 80 g + 45^(@)C + 20 g xx 30^(@)C)/( 80 g + 20 g) `
` = (( 80 xx 45)/30 + (20 xx 30)/100)^(@)C`
`= ( 36 + 6)^(@)C`
`= 42^(@)C`