Correct Answer - A
`A=1.7 W//m-^(@)C`
`(rho_w)=1000kg//m^(3)`
`L_(ice)=3.36xx10^(5) J/kg`
`L=10xx10^(-2) m`.
(a)`(Q)/(T)=(kA(theta_1-theta_2))/(l)`
implies .(Q)/(T)=(kA(theta_1-theta_2))/(Q)` ltbrge(kA(theta_1-theta_2))/(mL)`
`=(kA(theta_1-theta_2))/(Al rho_w L)`
`(1.7xx|(0-10)|)/(10xx1000xx3.36xx10^(5)xx10^(-2))`
`=(17xx10)/(3.36xx10^(7))`
`(17)/(3.36)=5.059xx10^(-7) m/sec`.ltbrge5xx10^(7)m/sec.`
(b) Let us assume taht x length of ice has been formed. To form a small strip of ice of length dx, dt time is required.
` (dQ)/(dt)= (KA(Delta theta ))/x`
`implies (dmL)/(dt)=(KA(Delta theta ))/x`
implies (Adx (rho_w) L)/(dt)=(KA(Delta theta))/x`
`implies (dx (rho_w) L)/(dt)=(K(Delta theta))/x`
implies dt=(xdx (rho_w)L)/(K(Delta theta)`
`implies int_(0)^(1)dt=((rho_w)L)/(K(Delta theta) int_(0)^(1)xdx`
`implies t=((rho_w)L)/(K(Delta theta)[(x^(2))/(2)]_(0)^(1)=((rho_w)L)/(K(Delta theta).1/2`
Putting values
`t=(1000xx3.36xx10^(5)xx(10xx10^(-2))/(1.4xx10xx2)`
`=(3.36)/(2xx17)xx10^(-2)xx10^(5)xx10^(3)xx10`
`=(3.36)/(2xx17)xx10^(7)sec.`
`=(3.36xx10^(7))/(2xx17xx3600) hrs.`
`=27.45 hr=27.5 hrs`.