Ice has formed on a shallow pond anda steady state has been reached with air abo

Question

Ice has formed on a shallow pond anda steady state has been reached with air above the ice at -5.20 C , and the bottom of the pond at 3.98 C. if the total depth of ice + water is 1.42, how thick is the ice? (Assume the thermal conductivites of ice and water are k=1.67 and k=0.502 W/m.K respectively)

Dont forget to use kelvins Please show all work, thanks! **On chegg they said the answer was 2 m, however the whole thing is 1.42 m only?** Plus if we use the equation K1T1/L1 + K2T2/L2 shouldnt it be equal to 273 and not 0 because we are dealing with kelvins?

Explanation / Answer

Heat flow = constant = -kA(T2-T1)/(x2- x1) since it is constant all across the sandwich we have
-k1A(T2-T1)/(x2- x1)= -k2A(T3-T2)/(x3- x2)
Where
(T1,x1) _____water_____(T2,x2) ___ice____(T3,x3)____air___
T1= 3.98C, T2=0C (from figure), T3=-5.2C
x3-x1=1.42m
x3-x2=?
k1 (T2-T1)/(x2- x1)= -k2(T3-T2)/(x3- x2)
k1 (T2-T1) (x3- x2)= k2(T3-T2)(x2- x1)

(x2-x1) / (x3-x2) = (k1/k2)[(T2-T1)/(T3-T2)]

Let (x2-x1)=a ; (x3-x2)=b
And we know that
a + b=1.42 and
a/b=(k1/k2)[(T2-T1)/(T3-T2)]=
a/b=(1.67/0.502)[(-3.98)/(-5.2)]=2.546
a= 2.546b substituting in a + b=1.42 we have

a + 2.546a =1.42
a= 0.4m or 40 cm (you can walk on that ice)
So
The water is 1.42 - 0.4 = 1.02m

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