Question

Can you please answer question 12 and maybe 13. Thank you.

An amount M; of ice at temperature Ti placed in thermal contact with a very large object which is maintained at Th. The thermal contact is via a rod L long with a cross sectional area A and a thermal conductivity k. The ice, the large object, and the rod, are otherwise thermally insulated r is 4200 K' an The specific heat for ice is 2100 steam is 2000 , and for The latent heat of fusion for the ice to water transition is 3.3 x 105J and the latent heat of vaporization for the water to steam or water kg K transition is 2.1 x 106 J 12. Given that Ti -50°, Th-400°, L -1.2m, A 3 x 10-2m2, and K, what is the rate at which the M 3kg of ice warms up? k 800w assume that the all the ice is at the same temperature (a) 0.7* 1.4 (c) 2.1K (d) 3.6K (e) 4.3K 13. Suppose that L were halved (so the new length is 0.5L) and the area were increased by 50% (so the new area is 1.5.4). What would the new rate of ice warming be? (a) 0.33 times the answer to the last question (b) 0.75 times the answer to the last question (c) 1.00 times the answer to the last question. d) 1.33 times the answer to the last question. (e) ***3.00 times the answer to the last question

Answer 1

Given that T_i = -50 degrees T_h = 400 degrees L = 1.2 m A = 3 times 10^-2 m^2 k = 800 W/mK we have to assume that the all the ice is at the same temperature. we know that: The heat transferred to the ice by heat conduction per second is given by delta Q/delta t = kA(T_h - T_i)/L = 800 times 3 times 10^-2(400 - (-50))/1.2 = 24(450)/1.2 = 10800/1.2 = 9000 J/s delta Q/delta t = 9000 J/s \r\n These entire heat goes into heating the mass m_i of ice so that the temperature rise of the ice per second i.e The rate at which the ice warms up can be found M_i c delta T = delta Q = 9000 where, c = specific heat capacity of ice. Therefore delta T = 9000/M_i c = 9000 times 3 times 2100 = 1.43 K/s delta T = 1.43 K/s Hence option (b) is correct.