A 10-cm bar is made from a 5-cm rod of copper fused together at the end with a 5

Question

A 10-cm bar is made from a 5-cm rod of copper fused together at the end with a 5-cm rod of aluminum. The cross-sectional area of the ends of the rods is 4cm2. The copper end of the rod is in thermal contact with steam at 100oC,while the aluminum end of the rod is in thermal contact with ice at 0oC. Neglecting heat losses from the sides of the bars, find:

a) The amount of ice in grams that melts in 10 minutes.

b) The amount of steam in grams that condenses in 10 minutes.

c) The temperature at the interface of the two rods.

Explanation / Answer

rate of heat transfer through conduction is given by

Q=k*A*temperature difference/d

where Q=heat transfer per second

k=thermal conductivity of material

A=cross section area

d=distance between two ends

thermal conductivity values :

copper=398 W/(m.degree celcius)

aluminium=237 W/(m.degree celcius)

as there is no heat loss, heat conducted through copper=heat conducted through aluminium

let temperature at interface of the two rods=T degree celcius

==>conductivity of copper*cross sectional area*temperature difference/length of copper rod

=conductivity of aluminium *cross section area *temperature difference/length of aluminium rod

as cross sectional area and length of each rod is equal,

398*(100-T)=237*(T-0)

==>398*100=398*T+237*T

==>T=62.677 degree celcius

then heat conducted through copper=heat conducted through aluminium

=398*4*10^(-4)*(100-62.677)/0.05=118.84 W

so in 10 minutes, heat lost by steam=118.84*10*60=71304 J

same amount of heat is being added to the ice.

part a:

latent heat of fusion of ice=334 kJ/kg

then mass melted by 71304 J of energy addition=71304/(334*1000)=0.21348 kg=213.48 grams

b)latent heat of condensation of steam is 2264.76 kJ/kg

then amount of steam condensed=energy lost/latent heat of condensation

=71304/(2264.76*1000)=0.031484 kg=31.484 grams

c)temperature at the interface is 62.677 degree celcius

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