Homework Help: Physics

Posted by sid on Tuesday, May 26, 2009 at 11:07pm.

A person makes a quantity of iced tea by mixing 500g of hot tea(c=4190J/(Kg.k) with an equal mass of ice(T=0c, L=333KJ/Kg). assume that mixture has negligible energy exchange with environment. If tea's initial temperature is 90c, then what is the temperature of mixture at the time of thermal equilibrim?

Physics - drwls, Wednesday, May 27, 2009 at 2:30am
The heat absorbed by the ice in melting and rising to equilibrium temperature equals the heat lost by the rest of the water. Let T be the final temperature.

0.5kg*4.190(90 - T) = 0.5kg*(L + 4.190T)

The 0.5 kg cancels out since the masses are equal. You know what L is; solve for T.

4.190(90-T) = 333 + 4190T
2*4.190T = 377.1 -333 = 44.1
T = 5.3C
Physics - sid, Wednesday, May 27, 2009 at 3:01am
thank u so much drwls

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