[Cotc]

hey wondering if anyone can help...

when 108g of water at a temp of 22.5 degrees C is mixed with 83g of water at an unknown temp, the final temp of resulting mixture is 47.9. what was the temp of the second sample of water? specific heat of water is 4.18 J deg-1 g-1.

this is what ive done:
q=s * m * T
  =4.18 * 108 * 22.5
  =10157.4

then for the second sample

T = 10157.4 / (83 * 4.18)
  = 29.277

i think that is the answer but im not sure if im misreading the question.

any help would be much appreciated.

[Donaldson Tan]

let's divide the water into 2 'sample'. the first sample is the 108g one and the second 'sample' is the 83g one.

let T be temperature of 2nd sample
heat gain by first 'sample' = m.c.dT = (108)(4.18)(47.9 - 22.5)
heat loss by second 'sample' = m.c.dT = 83(4.18)(T - 47.9)

assuming no heat loss to surroundings (and zero enthalpy of mixing),
heat gain by first 'sample' = heat loss by second 'sample'
(108)(4.18)(47.9 - 22.5) = 83(4.18)(T - 47.9)

T = 80.95 ~ 81^oC