Question

At 28.5°C, an aluminum ring has an inner diameter of 6.000 0 cm and a brass rod has a diameter of 6.070 0 cm. (Note, if needed, Table 1 is available for use in solving this problem.) (a) If only the ring is warmed, what temperature must it reach so that it will just slip over the rod? 486.11 This is the required temperature change of the ring. The problem asks for the final temperature. °C (b) If both the ring and the rod are warmed together, what temperature must they both reach so that the ring barely slips over the rod? 1988.63 X Be careful with round-off errors here. Keep all significant digits of intermediate results. °C (c) Would this latter process work? Yes No

Answer 1

T_i = 28.5 degree C d_i = 6.000 cm d_o = 6.070 If only Aluminium ring is warnel, temperature at which it slips over rod = ? That means find diameter of ring = diameter of rod Delta d = d_o alpha A_1 Dellta_T (6.070 - 6.000) = (6.000) (2.4 times 10^-5) Delta T Delta T = 486.11 T_f = 486.11+ T_i = 486.11 + 28.5 T_f = 514.611 degree C Now both ring a rod are warmed together. this means final diameter of both the ring & rod are equal so that one slip over the other. d_A1 = D_A_1 degree (1 + alpha_A1 Delta T) d_br = d degree_b (1+ alpha_br Delta T) d_Al = d_br rightwardsdoublearrow d degree_Delta l (1 + alpha_AL Delta T) = d degree_b (1 + alpha_br Delta T) 6.000(1 + 2.4 times 10^5 Delta T) = 6.070 (1 + 1.9 times 10^-5 (Delta T)) 1 + 2.4 times 10^-5 Delta T = 1.0116 (1 + 1.9 times 10^-5 Delta T) 2.4 times 10^-5 Delta T = 0.0116 + 1.22 times 10^-5 Delta T Delta T = 0.0116/0.4777 times 10^-5 Delta T = 2426.9 degree C T_f = 2426.9 + 28.5 = 2455.48 degree C = T_f