Asilver bar of length 30 cm and cross-sectional area 1.0 cm2 is used to transfer heat from a 100°c reservoir to a 0°c block of ice.

how much ice is melted per second? (for silver, k = 427 j/s⋅m⋅°c. for ice, lf = 334 000 j/kg.)
a. 4.2 g/s
b. 2.1 g/s
c. 0.80 g/s
d. 0.043 g/s

d. 0.043 g/s

Explanation:

Formula for rate of conduction of heat through a bar per unit time is as

follows

Q = k A ( t₁ - t₂ ) / L

A is cross sectional area and L is length of rod ,( t₁ - t₂ )  is temperature

difference . Q is heat conducted per unit time

Putting the values in the equation

Q = (427 x 1 x 10⁻⁴ x 100 )/ 30 x 10⁻²

= 14.23 J/s

mass of ice melted per second

= Q / Latent heat of ice

= 14.23 / 334000

= 0.043 g/s

total distance/total time

(85 + 82)/(5+3)

167/8

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