Question

Ice and COPPER STEEL water water A copper rod and a steel rod are joined to- gether as shown. Both have a cross sectional area of 4.0cm2. Both rods are insulated so that heat can be conducted only along their length. One end of the copper rod is immersed in boil- ing water, and one end of the steel rod is im- 65.0°C mersed in a mixture of ice and water 0°C. The temperature at the junction point is stable at 0 < 1.00m *-L2 65°C. [Since the temperatures are stable, this is an equilibrium situation. The rate of heat flow along the copper part of the rod equals the rate of heat flow along the steel rod.] (a) What is the rate of heat flow (in Watts) from the boiling water into the copper rod? (b) What is the rate of heat flow (in Watts) from the boiling water into the ice water? (c) Determine the length L2.

Answer 1

(a) The expression for the rate of heat flow is as follows: OPAT AL Here, p is the conductivity, A is the Area, L is the length, and AT is the temperature difference. The equation Qrate = P2 is rewritten as follows: AL Onde = P(T.-I) AL P(T; -T) AL Solve for Qrate - (58.14x10° S-m-^)(650-0°) rate= (4.0 cm) (1.0 m) (58.14x10 S-m-[(65+273.15 K)-273.15 K] (4.0 cmº) 1.0 cm(1.0 m) _(58.14x100 S-m--)[(338.15 K)–273.15 K] (4.0 cm? (10 (1.0m) (58.14x108 S-m-1)[65 K (4.0 cm) (106 m2 (1.0 m) Solving further for Qrata - (58.14x100 S- m')[65 K] Oster- (4.0 cm?) 10 cm (1.0m) =944. 7x1010 w

(b) At equilibrium, the heat current will be same, that is Qirate=944. 7x100 w Again PAT The equation Q. =- is rewritten as follows: Date = P(1, -1) AL P(T; -T) AL Rearrange the equation for L2. 1,- P(T; -T) 20-4 L2= Solve for L2 (58.14x100 S-m-') (65 K) (944.7x100 w)(4.0 cm²) (58.14x10 S-m-')(65 K) (944.7x10' w)(4.0 cm?) 106 mm = 1.00007 m 1.0 m Hence, the length is approx. 1.0 m.