Question

Two rods, one made of brass and the other made of copper, are joined end to end. The length of the brass section is 0.160 m and the length of the copper section is 0.640 m. Each segment has cross-sectional area 0.00680 m². The free end of the brass segment is in boiling water and the free end of the copper segment is in an ice and water mixture, in both cases under normal atmospheric pressure. The sides of the rods are insulated so there is no heat loss to the surroundings. What mass of ice is melted in 5.10 min by the heat conducted by the composite rod? [Hint: Thermal conductivities off brass and copper are 109.0 W/(m⋅K) and 385.0 W/(m⋅K), respectively.]

Answer 1

a) let temperature at the joint be T
then

K_c*A_c*(T-0)/(0.64) = K_b*A_b*(100-T)/(0.16)

where c is for copper and b is for brass.

Calculate to get T= 53.19 c

Heat transferred in 5.10 mins = 5.10*60*K_c*A_c*(53.19-0)/(0.64) = m_ice*L_{ice =} M_ICE* 333.55

mass of ice = 199.60 gram Answer

where L_ice= latent heat of melting of ice