Respuesta :
Given :
A copper rod has a length of 1.7 m and a cross-sectional area of [tex]3.2 \times 10^{-4} \ m^2[/tex].
One end of the rod is in contact with boiling water and the other with a mixture of ice and water.
To Find :
The mass of ice per second that melts.
Solution :
Amount of heat transfer from boiling water to ice through the copper bar is :
[tex]Q = \dfrac{kA\Delta T }{L}\\\\Q = \dfrac{390\times 3.2 \times 10^{-4} \times 100}{1.7}\\\\Q = 7.34 \ J/s[/tex]
Let, amount of ice melts per second is :
[tex]m=\dfrac{Q}{L_f}[/tex]
Here, [tex]L_f[/tex] is the latent of fusion of water and is equal to [tex]L_f = 3.34\times 10^5 \ J/s[/tex].
Putting all values in above equation, we get :
[tex]m=\dfrac{7.34}{3.34\times 10^5}\\\\m = 2.19 \times 10^{-5} \ kg/s[/tex]
Hence, this is the required solution.
