[zgns028sng]

 :-\213.5g of water were heated to a certain temperature and then mixed with 150.0g of water in a calorimeter.  The water in the calorimeter was at a temp of 23.6 degrees celcius.  The final temp of the mixture was 64.3 degrees celcius.  Find the original temp of the hot water.  Assume that the specific heat capacity of water is 4.184 / j/g degrees celcius

[sdekivit]

remember that the energy released by the hot water is taken up by the cold water and that you can equal them

just use:

213.5 * (T_initial - 64.3) = 150 * (64.3 - 23.6)

--> solve for T_initial