# 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//

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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.

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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`