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
Answer:
[tex]\frac{m}{t}=2.47*10^{-5} kg/s[/tex]
Explanation:
We can use the equation of the heat transfer
[tex]\frac{Q}{t}=\frac{kA\Delta T}{l}[/tex] (1)
And we know that Q can express in terms of rate change of mass and latent heat of fusion for the water Lf
[tex]\frac{Q}{t}=\frac{m}{t}*L_{f}[/tex] (2)
Here:
We can equal the equations 1 and 2 and solve it for m/t
[tex]\frac{m}{t}=\frac{kA\Delta T}{L_{f}*l}[/tex]
[tex]\frac{m}{t}=\frac{390*3.6*10^{-4}(100-0)}{3.34*10^{5}*1.7}[/tex]
[tex]\frac{m}{t}=2.47*10^{-5} kg/s[/tex]
I hope it helps you!
