One end of a cylindrical copper rod 1.2 m long with a 2 cm diameter is in boiling water at 100°C. The other end is in the ice at 0°C. b. How much time in seconds is required to melt 10 grams of ice?
1 Answer
Feb 1, 2017
Basic equation connecting thermal conductivity extrinsic factors like heat flow, temperature gradient area and length is
Note that
- Heat
#Q# required to melt#10# grams of ice#=mxxL#
where#L# is latent heat of fusion of water and is#=79.7cal cdot g^-1#
#Q=10xx79.7=797cal# - Thermal conductivity of Copper
#k=0.99 calcdot s^-1cm^-1·K^-1#
Inserting given values in equation (1) we get
#dot Q = 0.99 × (100-0) × (pixx1^2)/120 #
#=>dot Q = 2.59calcdot s^-1 # - Number of seconds required to conduct
#797cal# of heat#=797/2.59=308s# , rounded to last second