9. A long rod, insulated to prevent heat loss along its sides, is in perfect the

Question

9. A long rod, insulated to prevent heat loss along its sides, is in perfect thermal contact with boiling water (at atmospheric pressure) at one end and with an ice-water mixture at the other. The rod consists of a 1.00 m section of copper (one end in boiling water) joined end-to-end to a length L2 of steel (one end in the ice-water mixture). Both sections of the rod have cross-sectional areas of 4.00 cm2. The temperature of the copper-steel junction is 65.0 Degree C after a steady state has been set up (i.e., the heat transfer rate is the same through the copper and the steel). What is the length L2 of the steel section? The thermal conductivities for copper and steel are kc = 401 W/(m K) and k steel = 46 W/(m. K), respectively. 6cm 11cm 21cm 54cm 90cm

Explanation / Answer

Option (c) 21 cm is the correct answer.

Explanation:

Given,

L1 = 1 m; L2 = ?

A = 4 cm2 ; Temp of copper steel junction = T(c) = 65 deg C ; hot temp = T(h) = 100 deg cent of boiling water

k(Cu) = 401 W/(m.K) ; k(steel) = 46 W/(m.K)

We know that, the amount of heat flowing through the copper portion of rod will be given by:

H(Cu) = k(Cu) A [T(h) - T(c)] / L

H(Cu) = 401 W/(m.K) x 4 x 10-4 m2 [ 100 - 65] / 1 m = 5.614 W

We know that, H(steel) = H(Cu)

H(steel) = k(steel) A [ T(h) - T(c)] / L2

L2 =  k(steel) A [ T(h) - T(c)] / H(steel)

In this case, hot temp= T(h) = 65 deg and T(c) = 0 deg

L2 = 46 W/(m.K) x 4 x 10-4 m2 x ( 65 - 0 ) / 5.61 = 0.21 meters

Hence L2 = 0.21 m = 21 cm

Hence, option (c) 21 cm is the correct answer.

Related Questions

Navigate

• Browse (All)
• Browse 9
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.