Question

A copper rod has a length of 1.7 m and a cross-sectional area of 3.6 10-4 m2. One end of the rod is in contact with boiling water and the other with a mixture of ice and water. What is the mass of ice per second that melts

Answer 1

Answer:

$\frac{m}{t}=2.47*10^{-5} kg/s$

Explanation:

We can use the equation of the heat transfer

$\frac{Q}{t}=\frac{kA\Delta T}{l}$ (1)

And we know that Q can express in terms of rate change of mass and latent heat of fusion for the water Lf

$\frac{Q}{t}=\frac{m}{t}*L_{f}$ (2)

Here:

• k is the heat transfer coefficient of copper 390 J/s*m*C
• L(f) is the latent heat of fusion for the water 3.34*10⁵ J/kg
• ΔT is the difference in temperature 100 C boiling water and 0 C of ice

We can equal the equations 1 and 2 and solve it for m/t

$\frac{m}{t}=\frac{kA\Delta T}{L_{f}*l}$

$\frac{m}{t}=\frac{390*3.6*10^{-4}(100-0)}{3.34*10^{5}*1.7}$

$\frac{m}{t}=2.47*10^{-5} kg/s$

I hope it helps you!

Answer 2

Answer:

Mass of ice per second melt is 2.74×10^-5Kg/s

Explanation:

Temperature of one end of the copper rod is 100°C boiling point of water and the other end of the rod is 0°C

Temperature difference in the copper rod = 100 - 0 = 100°C

Cross sectional area = 3.6×10^-4m^2

Length of rod , L = 1.7m

Amount of heat transfer from the boiling water to the ice water mix through the copper rod is given by:

Q = KA◇T/ L

Q = (390×(3.6×10^-4)×100°C)/1.7

Q = 14.04/1.7

Q = 8.26J/s

From the equation

Q = mLf

m = Q/ Lf

Where Lf = Latent heat of fusion for water= 3.34×10^5J/Kg

m = 8.26/(3.34×10^5)

m = 2.74×10^-5Kg/s