When two bodies ofmasses `m_(1)` and `m_(2)` with specific heats `s_(1)` and `s_(2)` at absolute temperatures `T_(10)` and `T_(20)(T_(10) gt T_(20))` are connected by a rod of length l and cross sectional area A with thermal conductivity k. Find the temperature difference of the bodies after time t. Neglect any heat loss due to radiation at any surface..

Explanation

Let at time t after connecting the two bodies, they are at temperature, `T_(1)`and `T_(2)` respectively`(T_(1) gt T_(2))`. At this time, heat must be flowing from `m_(1)` to`m_(2)` as`m_(1)` is at higher temperature if dQ is the amount of heat flown through the rod from `m_(1)` to `m_(2)` and due to this if `dT_(1)` is fall in temperature of `m_(1)` and `dT_(2)` is rise in temperature of `m_(2)` then we have
`dQ = -m_(1)s_(1)dT_(1)` ..(4.17)
Also `dQ = m_(2)s_(2)dT_(2)` ....(4.18)
As this heat dQ is conducted through the rod , we have
`(dQ)/(dt) = (kA(T_(1))-T_(2))/(l)`
or `dQ = (kA)/(l) (T_(1)-T_(2))dt` ...(4.19) As in this expression dQ is given as a function of`(T_(1)-T_(2))`. Thus from (4.17)and (4.18),we can get
`dQ[(1)/(m_(1)s_(1))+(1)/(m_(2)s_(2))]=-d(T_(1)-T_(2))` ....(4.20)
Now from (4.19) and (4.20)
`(kA(T_(1)-T_(2)))/(l)[(1)/(m_(1)s_(1))+(1)/(m_(2)s_(2))]dt = -d(T_(1)-T_(2))`
or `(d(T_(1)-T_(2)))/(T_(1)-99T_(2))=(kA)/(l)[(1)/(m_(1)s_(1))+(1)/(m_(2)s_(2))]dt`
Integrating the expression within proper limits, we get
`int_(T_(10)-T_(20))^(T_(1)-T_(2))(d(T_(1)-T_(2)))/((T_(1)-T_(2))) = -int_(0)^(t) (kA)/(l)((1)/(m_(1)s_(1))+(1)/(m_(2)s_(2)))dt`
`In((T_(1)-T_(2)))/((T_(10)-T_(20)))=-(kA)/(l)((1)/(m_(1)s_(1))+(1)/(m_(2)s_(2)))dt`
or `(T_(1)-T_(2))=(T_(10)-T_(20))e^(-(kA)/(l))((1)/(m_(1)s_(2))+(1)/(m_(2)s_(2)))t`