posted by RoseBud on .
what mass of water at 10.0 degrees C is needed to cool 500.0ml of tea from 25.0 degrees C to 18.3 degrees C? Density and specific heat of the tea are the same as water.
This is done the same way as adding cool water to warm water. The only difference is that you solve for mass cool water rather than Tfinal.
(? x 4.184 x 6.7) = (500 x 4.184 x 8.3)
28.03x = 17363.6
x= 6.19 x 10(2)g
is this correct?
I don't think so.
It appears to me that you've reversed the delta T values.
[? x 4.184 x (8.3)] = 500 x 4.184 x 6.7
mass 10 C H2O needed approximately 400 g if I didn't punch the wrong keys.