10 g of water at `70^@C` is mixed with 5 g of water at `30^@C`. Find the temperature of the mixture in equilibrium. Specific heat of water is `1 cal//g.^@C`.

Explanation

Given, Mass m = 10g, m_(2) = 5g,
Temperature `T_(1)= 70^(@)`C, `T_(2) = 30^(@)`C
Let `T^(@)`C be the temperature of the mixture.
From principle of calorimetry, common temperature of the mixture at equilibrium is given by
`T= (m_(1)c_(w)T_(1) + m_(2)c_(w)T_(2))` = `(c_(w)(m_(1)T_(1) + m_(2)T_(2))/(c_(w)(m_(1)+m_(2))`
=`(m_(1)T_(1)+m_(2)T_(2))/(m_(1) + m_(2)) = (10xx70+5xx30)/(10+5) = (700+150)/(15)`
`=(850/15) = 56.67^(@)C`