Two rods with same length d and same sectional area A are made of copper (therma

Question

Two rods with same length d and same sectional area A are made of copper (thermal conductivity k_c) and aluminum (thermal conductivity k_a). The two rods are connected in series between two systems, one consisting of a mass m of ice at T_1 = 0 degree C (latent heat L), the other maintained at a constant temperature T_2 > T_1(see figure). Assume that there is no dissipation of heat to the surroundings. Calculate the temperature T_0 of the joint between the two rods. Calculate the heat transferred per unit time between the two systems. Calculate the mass of ice melting per unit time (dm/dt). Find the time necessary to melt all the ice. k_cT_1 + k_aT_2/()k_c + k_a). k_ck_a A/d (T_2 - T_1)/(k_c + k_a) 1/L k_c k_a A/d (T_2 - T_1)/(k_c + k_a). Lmd(k_c + k_a)/Ak_c k_a(T_2 - T_1).

Explanation / Answer

a) Here, m * To * (Kc + Ka) = m * Kc * T1 + m * Ka * T2

=> To * (Kc + Ka) = Kc * T1 + Ka * T2

=> temperature To of joint = (Kc * T1 + Ka * T2)/(Kc + Ka)

b)   Here, equivalent thermal conductivity = (KcKa)/(Kc + Ka)

=> Heat transferred = (A/d) * (T2 - T1) * [(KcKa)/(Kc + Ka)]

c) Here, dm/dt = (1/L) * Q

where, Q = heat transferred

=> mass of ice melting per unit time , dm/dt = (1/L) * (A/d) * (T2 - T1) * [(KcKa)/(Kc + Ka)]

d) time necessary = m * (1/dm/dt)

                              = m * Ld(Kc + Ka)/[AKcKa(T2 - T1)]

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