Question

If you have 370.0 mL of water at 25.00 °C and add 120.0 mL of water at 95.00 °C, what is the final temperature of the mixture? Use 1.00 g/mL as the density of water.

Answer 1

Answer on the question #55463 - Chemistry - General chemistry

Question:

If you have $370.0~\mathrm{mL}$ of water at $25.00^{\circ}\mathrm{C}$ and add $120.0~\mathrm{mL}$ of water at $95.00^{\circ}\mathrm{C}$, what is the final temperature of the mixture? Use $1.00~\mathrm{g/mL}$ as the density of water.

Solution:

Upon the mixing, the cold water is warming up and the hot water is cooling down. The exchange of the heat is summarized in a set of equations:

$$Q _ {c o l d} = m _ {c o l d} c \left(T _ {2} - T _ {1 - c o l d}\right)$$ $$Q _ {h o t} = m _ {h o t} c \left(T _ {2} - T _ {1 - h o t}\right)$$

where the amount of heat cold water gets and hot water gives are $Q_{cold}$ and $Q_{hot}$, respectively, $m_{cold}$ and $m_{hot}$ are the masses of water added, $c$ is water heat capacity and $T_{1}$ and $T_{2}$ are the initial and final temperature, respectively.

If we consider the law of conservation of energy, we find out that the $Q_{cold}$ is equal to $Q_{hot}$:

$$Q _ {c o l d} = - Q _ {h o t}$$ $$m _ {c o l d} c \left(T _ {2} - T _ {1 - c o l d}\right) = - m _ {h o t} c \left(T _ {2} - T _ {1 - h o t}\right)$$

Then, let's derive the final temperature $T_{2}$, taking into account assumption about $1\mathrm{g / mL}$ density of the water:

$$T _ {2} = \frac {m _ {c o l d} \times T _ {1 - c o l d} + m _ {h o t} \times T _ {1 - h o t}}{m _ {c o l d} + m _ {h o t}} = \frac {3 7 0 \times 1 \times 2 9 8 . 1 5 + 1 2 0 \times 1 \times 3 6 8 . 1 5}{3 7 0 \times 1 + 1 2 0 \times 1} = 3 1 5. 3 K$$

Answer: 315.3 K, or $42^{\circ}C$

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