A silver 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