Question

A hollow, iron sphere of inner radius 16 cm, outer radius 24 cm, and emissivity 0.45 has a 750 gram block of ice at 0◦C suspended at its center with a small drip pan underneath it.

(a) What must the temperature of the inside surface of the sphere be for the ice to completely melt in 35 seconds?

(b) What must the temperature at the outer surface of the sphere be for the inside temperature to be maintained during the process?

Answer 1

part a:

let temperature of the inside surface be T K. (T>273 K, where 273 K is melting point of ice)

then heat received by ice per second=emissivity*steffan's constant*area *(T^4-273^4)

=0.45*5.67*10^(-8)*pi*0.16^2*(T^4-273^4)

given that ice melts in 35 seconds.

energy required=mass*latent heat of fusion

=0.75*336 kJ=252 kJ

then 252*1000=0.45*5.67*10^(-8)*pi*0.16^2*(T^4-273^4)*35

==>T^4-273^4=3.5087*10^12

==>T=1369.1736 K=1096.1736 degree celcius

part b:

thermal conductivity of iron=79.5 W/(m.K)

area of the iron portion =pi*(0.24^2-0.16^2)=0.110531 m^2

if outer temperature is T degree celcius,

then heat transmitted through thermal conduction=heat being provided to ice

==>79.5*0.100531*(T-1096.1736)/0.08=0.45*5.67*10^(-8)*pi*0.16^2*(1369.1736^4-273^4)

==>T=1168.2437 degree celcius