Question

A 2.5kg of water at 65 degree celsius is added to a 45kg of water at 8 degree celsius. What is the final temperature of the mixture?

Answer 1

Answer on Question #72826 - Physics - Molecular Physics - Thermodynamics

Question:

A 2.5kg of water at 65 degree celsius is added to a 45kg of water at 8 degree celsius. What is the final temperature of the mixture?

Solution:

Let's assign the index 1 and 2 for the water with different temperature.

During the equilibration, the change in temperature occurs and it has to be included to the heat equation:

$$Q _ {1} = m _ {1} c \left(T _ {1} - T _ {x}\right)$$ $$Q _ {2} = m _ {2} c \left(T _ {x} - T _ {2}\right)$$

At the end point of equilibration, the heat of both systems becomes the same:

$$Q _ {1} = Q _ {2}$$ $$m _ {1} c \left(T _ {1} - T _ {x}\right) = m _ {2} c \left(T _ {x} - T _ {2}\right)$$

From this equation we can deviate the final temperature of the mixture:

$$T _ {x} = \frac {m _ {1} T _ {1} + m _ {2} T _ {2}}{m _ {1} + m _ {2}}$$ $$T _ {x} = \frac {2.5 \cdot 65 + 45 \cdot 8}{2.5 + 45} = 11{}^{\circ} \mathrm{C}$$

Answer: The final temperature of water mixture is $11{}^{\circ}\mathrm{C}$.

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