Question

Suppose you mix 7.00 L of water at $2.00\cdot {10}^1{}^{\circ }\mathrm{C}$ with $3.00\mathrm{L}$ of water at ${32.0}^{\circ }\mathrm{C}$; the water is insulated so that no energy can flow into it or out of it. (You can achieve this, approximately, by mixing the two fluids in a foam cooler of the kind used to keep drinks cool for picnics.) The $10.0\mathrm{L}$ of water will come to some final temperature. What is this final temperature?

Short answer

Answer: The final temperature of the 10.0 L of water after mixing is 23.6°C.

Step-by-step solution

Calculate the mass of each water sample

First, we need to calculate the mass of each water sample. Since the density of water is approximately 1 g/mL (or 1 kg/L), we can directly multiply the volume of each water sample by 1 kg/L to get the mass. Mass of the first water sample (m1): $m_1=7.00\mathrm{L}\times 1\mathrm{kg}/\mathrm{L}=7.00\mathrm{kg}$ Mass of the second water sample (m2): $m_2=3.00\mathrm{L}\times 1\mathrm{kg}/\mathrm{L}=3.00\mathrm{kg}$

Set up the heat transfer equation

Next, we need to set up the heat transfer equation. Since the heat gained by one body (Q1) is equal to the heat lost by the other body (Q2) in an insulated system, we can write the equation as: $Q_1=Q_2$ When a substance gains or loses heat, we can calculate the heat transfer (Q) using the formula: $Q=mc(T_f-T_i)$ Where: - Q is the heat transfer - m is the mass of the substance - c is the specific heat of the substance - T_f is the final temperature - T_i is the initial temperature Since the specific heat of water (c) is the same for both samples, we can write the heat transfer equation as: $m_{1}c(T_f-T_{i1})=m_{2}c(T_{i2}-T_f)$

Solve for the final temperature (Tf)

Now we have the equation: $7.00\times 4.186\times {10}^3(T_f-20.0)=3.00\times 4.186\times {10}^3(32.0-T_f)$ First, we can simplify the equation by dividing both sides by $4.186\times {10}^3$: $7.00(T_f-20.0)=3.00(32.0-T_f)$ Now, distribute the coefficients: $7.00T_f-140.0=96.0-3.00T_f$ Move all terms related to T_f to one side and constants to the other: $7.00T_f+3.00T_f=96.0+140.0$ Combine like terms: $10.00T_f=236.0$ Finally, divide both sides by 10.00 to find T_f: $T_f={23.6}^{\circ }\mathrm{C}$

Conclusion

The final temperature of the 10.0 L of water after mixing the two samples is 23.6°C.