Joule’s paddle-wheel experiment (energy balance) — An insulated container holds 20 kg of water at 25°C. An agitator is driven by a 40 kg mass falling through 4 m, and this is repeated 500 times. Take g = 9.8 m/s² and neglect agitator heat capacity. The final water temperature (°C) will be:

Introduction:
Joule’s classic experiment demonstrated equivalence between mechanical work and heat. Here, potential energy from a falling mass is converted to internal energy (sensible heat) of water in an adiabatic container, raising its temperature.

Given Data / Assumptions:

• Water mass m_w = 20 kg; initial temperature T_i = 25°C.
• Falling mass m = 40 kg; drop height h = 4 m; repetitions n = 500.
• Acceleration due to gravity g = 9.8 m/s².
• Container insulated; no heat loss; c_p,water ≈ 4.186 kJ/kg·°C.

Concept / Approach:
Total mechanical work becomes sensible heat of the water: Q = m_w * c_p * ΔT. The work per drop is W_drop = m * g * h. The total work is n * W_drop.

Step-by-Step Solution:
Compute work per drop: W_drop = 40 * 9.8 * 4 = 1568 J.Total work: W_total = 1568 * 500 = 784000 J.Heat capacity of water: m_w * c_p = 20 kg * 4186 J/kg·°C = 83720 J/°C.Temperature rise: ΔT = W_total / (m_w * c_p) = 784000 / 83720 ≈ 9.365°C.Final temperature: T_f = 25 + 9.365 ≈ 34.4°C.

Verification / Alternative check:
Unit consistency: J divided by J/°C gives °C rise. Neglecting losses and agitator heat capacity is standard for this idealized problem, matching textbook answers.

Why Other Options Are Wrong:

• 40.5°C and 31.6°C imply different energy inputs or water masses.
• 26.8°C suggests minimal work input; incorrect given the calculated ΔT.
• 25.0°C would require zero net work or complete heat loss, contrary to assumptions.

Common Pitfalls:
Using c_p = 4.2 kJ/kg·°C vs. 4.186 kJ/kg·°C changes T_f only slightly; misplacing decimal in total work or repetitions causes large errors.

Final Answer:
34.4