Question

A lagged stick of cross section area 1 cm and length 1m is initially at a temperature of 0°C. It is then kept
between 2 reservoirs of temperature 100°C and 0°C. Specific heat capacity is 10 J/kg°C and linear mass
density is 2kg/m. Find total heat absorbed by the rod to reach steady state (in kJ)
100°C
oc
-
oºo​

Answer 1

Answer:

Explanation:

It is given that,

Specific heat = 10 j / kg°C

linear mass  density = 2kg/m.

Tempereture gradient = dT / dx

dT / dx = T - 100 / x - 0

= -100° C/m

T = 100 (1 - x) °C

Total heat absorbed = ∫ mass * specific heat * change of temperature

= $\int\limits^1_0 {2dx}$*10*(T-0)

= 20 $\int\limits^1_0 {100(1-x)dx}$

= 2000 [x - x²/2]²₀

= 1000 j

Thus, total heat absorbed by the rod to reach steady state is 1000 j.