The two ends of a metal rod are maintained at temperatures 100°C  and 110° C. The rate of heat flow in the rod is found to be 4.0 J/s. If the ends are maintained at temperatures 200°C and 210°C, the rate of heat flow will be

Rate of heat flow through a rod is given by dQdt=-KAdTdx Let length of the rod be L .  Case I : dTdx=ΔTΔx=110-100L=10L∴  dQ1dt=-KA10L                                            . . . .(i) Also, dQ1dt=4 J s-1                                           . . . .(ii) Case II : dTdx=ΔTΔx=210-200L=10L∴  dQ2dt=-KA10L                                               . . . .(iii) So, from equations (i), (ii) and (iii) dQ2dt=dQ1dt=4 J s-1