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A small vessel holds a mass m of water at initial temperature theta1 and is connected by a metal rod (length L, cross-sectional area A, thermal conductivity K) to a large tank kept at a constant higher temperature theta0 (theta1 < theta0). The rod is laterally insulated, the water has specific heat capacity s, and all other heat capacities are negligible. Find the time taken for the water in the small vessel to warm up to theta2, where theta1 < theta2 < theta0.

1. t = (m*s*L)/(K*A) * ln((theta0 - theta1)/(theta0 - theta2))
2. t = (K*A)/(m*s*L) * ln((theta0 - theta1)/(theta0 - theta2))
3. t = (m*s*L)/(K*A) * ln((theta0 - theta2)/(theta0 - theta1))
4. t = (m*s*L)/(K*A) * (theta2 - theta1)/(theta0 - theta1)

Correct answer: t = (m*s*L)/(K*A) * ln((theta0 - theta1)/(theta0 - theta2))

Solution

Heat flows through the rod at rate dQ/dt = K*A*(theta0 - theta)/L, where theta is the instantaneous water temperature. This heat raises the water temperature: m*s*(d theta/dt) = K*A*(theta0 - theta)/L. Separating variables and integrating produces an exponential approach of theta toward theta0, analogous to Newton's law of cooling. Solving for the time to reach theta2 gives a logarithm of the ratio of the initial to the remaining temperature difference.

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