posted by sid .
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?
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
thank u so much drwls