Answer :
The principle of conservation of energy and assuming no heat loss to the surroundings, the final temperature of the water mixture is approximately 100.28°C.
Step 1: Convert the volumes of water to mass. Since the density of water is 1g/mL, the mass of the hot water is 240g and the mass of the cold water is 60g.
Step 2: Determine the heat gained or lost by each sample of water using the equation:
q = m * c * ΔT
where q is the heat gained or lost, m is the mass of the water, c is the specific heat capacity of water (4.18 J/g°C), and ΔT is the change in temperature.
For the hot water:
[tex]q_hot = m_hot * c * ΔT_hot[/tex]
[tex]q_hot = 240g * 4.18 J/g°C * (T_final - 72.8°C)[/tex]
For the cold water:
[tex]q_cold = m_cold * c * ΔT_cold[/tex]
[tex]q_cold = 60g * 4.18 J/g°C * (T_final - (-3.5°C))[/tex]
Step 3: Since the heat gained by the hot water is equal to the heat lost by the cold water (assuming no heat loss to the surroundings), we can equate the two equations from Step 2:
[tex]240g * 4.18 J/g°C * (T_final - 72.8°C) = 60g * 4.18 J/g°C * (T_final - (-3.5°C))[/tex]
Step 4: Simplify and solve for T_final.
[tex]960.96 * (T_final - 72.8) = 250.8 * (T_final + 3.5)[/tex]
Step 5: Expand and rearrange the equation to isolate T_final:
[tex]960.96 * T_final - 70380.288 = 250.8 * T_final + 877.8[/tex]
[tex]710.16 * T_final = 71258.088[/tex]
[tex]T_final = 71258.088 / 710.16[/tex]
[tex]T_final ≈ 100.28°C[/tex]
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