Two blocks of the same metal and mass are at different initial temperatures T1 a

Question

Two blocks of the same metal and mass are at different initial temperatures T1 and T2. The blocks are brought into contact and come to a final temperature Tf. Assume the system is totally isolated from the surroundings and that the surroundings are at equilibrium. a. Your intuition probably tells you that for 2 identical blocks, Tf is the average of the initial temperatures, so that Tf = 1/2 (T1 + T2). Show that for a system of two blocks totally isolated from the surroundings that this is true. b. Show that the change in entropy is ?S = CP ln T1 + T2 ( )2 4T1 T2 c. How does this expression show that this process is spontaneous?

Explanation / Answer

At equilibrium, the two blocks will have the same temperature (this turns out to be the state that corresponds to the maximum entropy). Because this process occurs in a thermally isolated system, energy is conserved, and the heat energy gained by the initially cool block is equal to the heat lost by the initially hot block: The specific heat is defined as C = (1/m)*dq/dT, where m is a unit mass of material, T is the temperature, and q is the heat energy, so dq = m*C*dT If we assume the specific heat is constant (doesn't depend on temperature), this integrates to: ?q = m*C*?T In this question, both blocks are the same material and C is the same for both, so we have: ?q_total = 0 = ?q_c + ?q_h = C*(m_c*(T_eq - T_c) + m_h*(T_eq - T_h)) where ?q_c, m_c, and T_c are the change in heat, mass, and initial temperature of the cold block, and ?q_h, m_h, and T_h are the corresponding parameters for the hot block. T_eq is the equilibrium temperature. Solving for T_eq gives: T_eq = (m_c*T_c + c_h*T_h)/(m_c + m_h) The equilibrium temperature is just the mass-weighted average of the temperatures of the two blocks. (Note that if the blocks were made of different materials, the average would also be weighted by the specific heats of the materials.) Plugin the numbers for this problem

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