A cylindrical block of length 0.4 m and area of cross-section `0.04 m^2` is placed coaxially on a thin metal disc of mass 0.4 kg and of the same cross - section. The upper face of the cylinder is maintained at a constant temperature of 400 K and the initial temperature of the disc is 300K. if the thermal conductivity of the material of the cylinder is `10 "watt"// m.K ` and the specific heat of the material of the disc is `600J//kg.K`, how long will it take for the temperature of the disc to increase to 350 K? Assume for purpose of calculation the thermal conductivity of the disc to be very high and the system to be thermally insulated except for the upper face of the cylinder.

Answer

166.36

Explanation

For a change in temperature `d theta` of disc of mass M, heat given `dQ=Msdtheta` of where S is specific heat.
Thus the rate of heat supplied `(dQ)/(dt)=MS(d theta)/(dt)` Also, for the cylindrical block of cross-sectional area length , maintained at one end at temperature of `theta_(2)` . The rate of heat follow for the other end at temperature `theta` is
`(dQ)/(dt)=KA((theta_(2)-theta))/(1)or(d theta)/(theta_(2)-theta)=(KA)/(MSI)dt`
Integrating the equation
`int_(theta_(i))^(theta_(t))(d theta)/((theta_(2)-theta))=(KA)/(MSI)int_(t_(1))^(t_(2))dt`
`t_(2)-t_(1)=(MSI)/(KA)In((theta_(2)-theta_(1))/(theta_(2)-theta_(1)))`
Gives `(t_(2)-t_(1))=(MSI)/(KA)ln[(400-300)/(400-350)]=240ln2=166.36sec`.