If you add two 45g ice cubes at 0C to some tea (500mL) at 20C. Calculate the fin

Question

If you add two 45g ice cubes at 0C to some tea (500mL) at 20C. Calculate the final temp at equilibrium. For tea, use same constants as H20.

Ok, I know you find the energy of Ice by 333j/g X 90 g ice and after that I think you set that Joule equal to that of (mass water)(4.184)(change in temp.) and sole for the change in temp. Then you set the two new (mass)(Specific heat)(T_final-T_initial)=-(mass Ice water)(4.184)(T_final-T_Initial).

Or something like that... keep getting wrong answer though. :/

Explanation / Answer

heat loss by water = heat gained by ice

500 x 4.184 x (20-T) = 90 x 333.5+ 90 x4.184 x ( T-0)

41840 -2092T = 30015 + 376.56T

T = 4.79 C

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