Question

A mass 2.4 kg of solid laurasium is initially at temperature 6∘C. It is put into contact with 3.8 kg of liquid tethium which is at temperature 97∘C, and the two chemicals are thermally insulated from the rest of the universe. The specific heat capacity of solid laurasium is 1165Jkg∘C, and the specific heat capacity of liquid laurasium is 3495Jkg∘C. The melting point of laurasium is 22∘C, and the latent heat for its solid liquid transition is 25149Jkg. The specific heat capacity of liquid tethium is 2330Jkg∘C. This material does not undergo a phase transition in this problem. What is the final equilibrium temperature of the laurasium and the tethium? Answer in degrees Celsius. Ans. 54.4

Answer 1

Solution we have put 214 kg of solid laurasium into a 3.8 kg of liquid te thium at gi°c. heat lost by liquid te thium heat gain by laurasium mq AT = mecera Tot me to the ce ate mt = 3.8 where, where, 4 7 2330 I likg. oc 3.8 kg and AT [97°– TA] Tf » final equilibrium temp. and me • 4 kg, Ces 1165 J 1kg. cc. and AT 22.6 - 6a 16.6. and Le 25149 Ilkg x che = 3495 Tlk.oC ste [TB - 22°C] = 2.4 Norco put all these values in equation ④ 3.8 x 2 330 (97-T;) = 2.4*116 5 x 16 + 2.4 X 25,149 + 2.4 x 349 5x [ T1 - 22°c.] 85%,838 – 8,854 TF 44736_t 60,357.6. +8,388 TH 184,536 938,280.4 17 242 Tf ) It 938, 280.4 17 242 IM TE 54.4°C Ave.