# If 200 Grams Of Water At 85 Degrees Celsius Were Added To 150 Grams Of Water At 15 Degrees Celsius, What Would Be The Final Equilibrium Temperature Of The Mixture?

This is about thermal physics, not a chemical question at all.

If there's no energy lost from the system then,
energy lost by hot substance cooling down = energy gained by cold substance heating up.
Where
mc(T(hot) - T(equilibrium) ) = mc (T(equilibrium) - T(cold))
where
m = mass
c = specific heat capacity
T = temperature

since c are the same
200*(85-x)=150(x-15)
4*(85-x)=3*(x-15)
340-4x=3x-45
then
7x=385
x=55 Celsius
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Two q's to 0 problem and a common chemistry exercise.

q(joules) = mass * specific heat * change in temperature Celsius

q cancels algebraically, of course, so....

(200 g H2O)(4.180 J/gC)(Tf - 85 C) + (150 g H2O)(4.180 J/gC)(Tf- 15 C) = 0

(836Tf - 71,060) + (627Tf - 9,405) = 0

1463Tf = 80,465

Temperature final = 55 degrees Celsius

======The answer you have but this is how this problem would be done in chemistry.

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