Q. 18

A metal rod of cross-sectional area 1.0 cm3 is being heated at one end. At one time, the temperature gradient is 5.0°C cm–1 at cross-section A and is 2.5°C cm–3 at cross-section B. calculate the rate at which the temperature is increasing in the part AB of the rod. The heat capacity of the part AB = 0.40 J°C–1, thermal conductivity of the material of the rod = 200 W m–1 °C–1. Neglect any loss of heat to the atmosphere.

Answer :


Given:
Cross sectional area of the metal rod: A = 1 cm2= 0.0001 m2
Temperature gradient at A :
(dT/dx)A = 5.0 °C cm–1= 500 °Cm-1.
Temperature gradient at B : (dT/dx)B = 250 °C m–1.
Heat capacity of the rod AB : C = 0.40 J°C–1.
Thermal conductivity of the material of the rod :
K = 200 W m–1 °C–1.
Formula used:
Rate of amount of heat flowing or heat current is given as:

Here, Δθ is the amount of heat transferred, ΔT is the temperature difference, K is the thermal conductivity of the material, A is the area of cross section of the material and x is the thickness or length of the material.
Rate of heat flow is dθ/dt
Hence at cross section A:



At cross section B:



Now, the rate of flow of heat throughout the rod AB is



Δθ= Q = msΔT
here, Q is the amount of heat, m is the mass of the material, s is the specific heat of the material and ΔT is the change in temperature.
And Heat Capacity is : C = ms = Q/ΔT

Substituting we get,



Hence, the rate at which temperature increases is 12.5 °C /s.


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