# 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?

## 2 Answers

Sosuke Okamura answered
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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John McCann answered

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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