Question

Suppose you have a 0.3 kg icicle (c=2093 J/kg K) at -8 C. You want to eventually turn this chunk into steam (c= 1890 J/kg K) at a temperature of 110 C. The only thing you possess which will help you is an easy-bake oven. It is powered by a 60 W light bulb which is only 70% efficient. How long will it take to achieve turning the icicle into steam? (L_fus =3.33*10⁵ J/kg, L_vap =2.26*10⁶ J/kg)

Answer 1

Given

mass m = 0.3 kg , temperature T1 = -8 0C

specific heat of ice C_i = 2093 J/kg.k,Lfus =3.33*10^5 J/kg, Lvap =2.26*10^6 J/kg

stem at temperature is 110 0C

here the ice should convert in to water it happens at 0 0c ,then water should turn in to steam at 100 0c , later should reach to 110 0c

the equation can be written as

total heat energy required is  

Q = m(C_i(0-(-8))+L_fus+C_w(100-0)+L_vap+C_steam(110-100))

susbtituting the values

Q = 0.3(2093*8+3.33*10^5+4186(100)+2.26*10^6 +1890(10)) J

Q = 914173.2 J

easy bake oven of 60 w and 70% efficiency is it can supply of 42 J of energy per second

so time taken to supply the energy of 914173.2 J energy is

t = 914173.2/42 s = 21766.0 s = 362.77 min = 6.04 hr